Question 1

For what value of n will the nth term of the arithmetic progression 15, 12, 9, ... be equal to the nth term of the arithmetic progression -15, -13, -11, ... ?

Accepted Answer

Introduction / Context: This question checks understanding of arithmetic progressions and the formula for the nth term. Two different arithmetic progressions are given, and we are asked to find the value of n for which their nth terms are equal. It tests algebraic manipulation and careful handling of signs. Given Data / Assumptions: First arithmetic progression: 15, 12, 9, ... Second arithmetic progression: -15, -13, -11, ... The nth term of the first sequence should equal the nth term of the second sequence. We assume n is a positive integer. Concept / Approach: The nth term of an arithmetic progression with first term a and common difference d is given by a + (n - 1) * d. We find nth term expressions for both sequences and then set them equal. Solving the resulting linear equation in n gives the point where both sequences share the same term value. Step-by-Step Solution: Step 1: For the first progression 15, 12, 9, ... the first term a1 is 15 and the common difference d1 is 12 - 15 = -3.Step 2: The nth term of the first progression is T1(n) = 15 + (n - 1) * (-3) = 15 - 3(n - 1).Step 3: Simplify T1(n): 15 - 3(n - 1) = 15 - 3n + 3 = 18 - 3n.Step 4: For the second progression -15, -13, -11, ... the first term a2 is -15 and the common difference d2 is -13 - (-15) = 2.Step 5: The nth term of the second progression is T2(n) = -15 + (n - 1) * 2 = -15 + 2n - 2 = 2n - 17.Step 6: Set the nth terms equal: 18 - 3n = 2n - 17.Step 7: Solve for n: 18 + 17 = 2n + 3n, so 35 = 5n and n = 35 / 5 = 7. Verification / Alternative check: Compute the 7th term of each sequence and compare. For the first progression, T1(7) = 18 - 3 * 7 = 18 - 21 = -3. For the second progression, T2(7) = 2 * 7 - 17 = 14 - 17 = -3. Both give the same value, confirming that n = 7 is the correct solution. Checking smaller n values such as 1 or 2 quickly shows that the terms are not equal for those. Why Other Options Are Wrong: n = 1: First term of first progression is 15, and of second is -15, not equal. n = 2: Second terms are 12 and -13, which are different. n = 5: Fifth terms are different and do not match -3. Common Pitfalls: Errors often arise from incorrect computation of the common difference or from mistakes when simplifying the nth term formula. Some students also forget to distribute the negative sign correctly in 15 - 3(n - 1). Others mistakenly equate a1 + nd instead of a1 + (n - 1)d, which shifts the sequence and leads to a wrong answer. Keeping the standard form of the nth term clear avoids these issues. Final Answer: The nth terms of the two arithmetic progressions are equal when n = 7.

Question 2

In how many different ways can 4 girls and 5 boys be arranged in a single row so that all 4 girls always stand together as one consecutive group?

Accepted Answer

Introduction / Context: This question is a classic arrangement problem in permutations where a subset of people must always stand together. The 4 girls must be treated as a single block, and within that block they can be permuted among themselves. This checks understanding of grouping in permutations. Given Data / Assumptions: There are 4 distinct girls and 5 distinct boys. We arrange all 9 people in a single row. All 4 girls must stand together as one consecutive group. Different orders of the same individuals count as different arrangements. Concept / Approach: We first treat the 4 girls as a single super person or block. Then we count permutations of this block together with the 5 boys. After that, we account for the internal permutations of the 4 girls inside the block. The total arrangements are the product of these two counts. Step-by-Step Solution: Step 1: Consider the 4 girls as a single block G.Step 2: We now have 5 boys plus the one block G, making 6 entities in total.Step 3: The number of ways to arrange these 6 entities in a row is 6! (because all are distinct at this level).Step 4: Compute 6! = 720.Step 5: Inside the girls block, the 4 distinct girls can be arranged among themselves in 4! ways.Step 6: Compute 4! = 24.Step 7: Combine the counts: total arrangements = 6! * 4! = 720 * 24 = 17280. Verification / Alternative check: Why Other Options Are Wrong: 18000, 17829 and 18270: These values do not match the exact product of 6! and 4!, and likely result from arithmetic mistakes or miscounting entities such as using 7! or only 5! instead of 6!. Common Pitfalls: Many students forget to multiply by the internal permutations of the 4 girls and stop at 6!, which undercounts. Others mistakenly treat girls and boys as identical groups or miscount the total number of entities to arrange as 9 instead of 6 after grouping. It is important to clearly separate the outer arrangement of blocks and individuals from the internal arrangement within the block. Final Answer: The number of ways to arrange the 4 girls and 5 boys with all girls together is 17280.

Question 3

In how many different ways can all the letters of the word POVERTY be arranged to form distinct letter sequences?

Accepted Answer

Introduction / Context: This question is a straightforward permutation problem where all letters in a word are distinct. We are asked to count the number of different sequences that can be formed using all letters of the word POVERTY, without any additional conditions. Given Data / Assumptions: The word is POVERTY. Letters in POVERTY are P, O, V, E, R, T, Y. No letter is repeated; all 7 letters are distinct. We must use all letters in each arrangement. Different orders of the same letters count as different arrangements. Concept / Approach: When all letters are distinct and we use all of them, the number of arrangements equals the factorial of the number of letters. Since there are 7 distinct letters, the count of permutations is 7!. Step-by-Step Solution: Step 1: Count the number of distinct letters in POVERTY. There are 7 distinct letters.Step 2: To form an arrangement using all letters, we are essentially permuting 7 distinct items.Step 3: The number of permutations of 7 distinct items is 7!.Step 4: Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1.Step 5: Calculate step by step: 7 * 6 = 42, 42 * 5 = 210, 210 * 4 = 840, 840 * 3 = 2520, 2520 * 2 = 5040.Step 6: Therefore, there are 5040 distinct arrangements. Verification / Alternative check: Why Other Options Are Wrong: 2520: This equals 7! / 2 and would be correct only if one letter were repeated twice, which is not the case here. 1260: This is 7! / 4 and does not correspond to any standard repetition structure in this word. None: The correct count is already represented by option 5040, so None is incorrect. Common Pitfalls: Students sometimes mistakenly think some letters repeat, for example confusing POVERTY with words that have double letters, and use formulas for permutations with repetitions. Another pitfall is arithmetic error when computing 7!. Writing out the factorial multiplication step by step and double checking helps avoid such mistakes. Final Answer: The number of different arrangements of the letters of POVERTY is 5040.

Question 4

In how many different ways can all the letters of the word ABYSMAL be arranged, given that the letter A appears twice and all other letters are distinct?

Accepted Answer

Introduction / Context: This question asks for the number of permutations of the letters of the word ABYSMAL. Here, one letter is repeated, which means we must adjust the basic factorial formula for permutations of distinct items to account for indistinguishable letters. Given Data / Assumptions: The word is ABYSMAL. Letters are A, B, Y, S, M, A, L. The letter A appears twice; all other letters (B, Y, S, M, L) appear exactly once. We must use all letters in each arrangement. Arrangements that differ only by swapping the two As are not considered different. Concept / Approach: If there are n total letters, with one letter repeated r times and the rest distinct, then the number of distinct permutations is n! / r!. This divides out the overcount due to the indistinguishability of the repeated letters. Here we have 7 letters in total with A repeated 2 times, so we use 7! / 2!. Step-by-Step Solution: Step 1: Count total letters in ABYSMAL. There are 7 letters.Step 2: Count how many times each letter appears. A appears 2 times; B, Y, S, M, L appear 1 time each.Step 3: Use the formula for permutations with one repeated letter: total permutations = 7! / 2!.Step 4: Compute 7! = 5040.Step 5: Compute 2! = 2.Step 6: Divide: 5040 / 2 = 2520. Verification / Alternative check: Why Other Options Are Wrong: 5040: This would be 7! and treats both As as distinct, overcounting arrangements where As are swapped. 3650 and 4150: These values do not correspond to any simple factorial ratio and likely arise from arithmetic guesses rather than correct combinatorial reasoning. Common Pitfalls: A common mistake is to forget that the two As are identical and simply use 7!, which is wrong. Some students also incorrectly divide by more than 2!, imagining extra repetitions that do not exist. Carefully counting how many times each letter appears and then applying the n! divided by product of factorials of repeated counts is essential. Final Answer: The total number of different arrangements of the letters of ABYSMAL is 2520.

Question 5

Seven students are to be allotted seven beds in a hostel hall. Two of them, Parvin and Anju, refuse to take adjacent beds because Anju snores. In how many different ways can all seven students be allotted the seven beds so that Parvin's bed is not n…

Accepted Answer

Introduction / Context: This arrangement problem adds a restriction of non adjacency for two specific students. We must count permutations of seven distinct students with the condition that Parvin and Anju are not seated or allotted beds next to each other. It demonstrates the use of complementary counting in permutations. Given Data / Assumptions: There are 7 distinct students. There are 7 distinct beds in a line. Parvin and Anju must not occupy adjacent beds. Any allotment is a one to one assignment of students to beds. Concept / Approach: Instead of directly counting the valid arrangements, it is easier to use the complement. First count all possible arrangements of the 7 students without restrictions. Then subtract arrangements in which Parvin and Anju are adjacent. The remaining arrangements are exactly those where Parvin and Anju are not next to each other. Step-by-Step Solution: Step 1: Total unrestricted arrangements of 7 distinct students in 7 distinct beds is 7! = 5040.Step 2: To count arrangements where Parvin and Anju are adjacent, treat them as a single block PA or AP.Step 3: The block (PA or AP) plus the other 5 students makes 6 entities to arrange.Step 4: The number of ways to arrange these 6 entities in a line is 6! = 720.Step 5: Inside the block, Parvin and Anju can switch places in 2 ways: PA or AP.Step 6: Therefore, total arrangements where they are adjacent = 6! * 2 = 720 * 2 = 1440.Step 7: Now subtract these from the total: valid arrangements = 5040 - 1440 = 3600. Verification / Alternative check: Why Other Options Are Wrong: 2400 and 6400: These do not align with 7! minus a reasonable multiple of 6! and likely result from miscounting the adjacency cases or forgetting the internal arrangements of the block. 7200: This is larger than the total number of possible arrangements (5040) and so cannot be correct. Common Pitfalls: Some learners count the adjacent cases incorrectly by treating the block as a single entity but forgetting to multiply by 2 for the internal arrangement of Parvin and Anju. Others mistakenly use 5! instead of 6! for the outside arrangements. Another frequent error is to forget to subtract from the total and directly use the count of adjacent cases as the answer. Final Answer: The number of ways to allot the beds so that Parvin and Anju are not next to each other is 3600.

Question 6

If the letters of the word CHASM are rearranged in all possible different 5 letter words and these words are listed in dictionary (lexicographic) order, what is the rank (position) of the word CHASM in this ordered list?

Accepted Answer

Introduction / Context: This question is about finding the lexicographic rank of a word formed from distinct letters. You must imagine all permutations of the letters arranged in dictionary order and determine the position of a given word in that list. This is a typical application of permutations with positional counting. Given Data / Assumptions: The word is CHASM. Letters are C, H, A, S, M, all distinct. All permutations of these 5 letters are considered. The words are sorted alphabetically as in a dictionary. We want the rank of CHASM among these permutations. Concept / Approach: We calculate how many words would appear before CHASM when all permutations are listed in dictionary order. For each position from left to right, we count permutations that start with a smaller letter than the current one, given the remaining letters. The final rank is one plus the total number of words that would appear before the target word. Step-by-Step Solution: Step 1: List the letters of CHASM in alphabetical order: A, C, H, M, S.Step 2: Consider the first position. The first letter of CHASM is C. Letters smaller than C in the sorted list are only A.Step 3: For all words starting with A, the remaining 4 letters can be arranged in 4! ways, which is 24. These 24 words come before any word starting with C.Step 4: Now fix the first letter as C. Remaining letters are H, A, S, M. Sort them: A, H, M, S. The second letter in CHASM is H.Step 5: Letters smaller than H among the remaining ones are only A. If second letter is A, the remaining 3 letters can be arranged in 3! = 6 ways. So 6 more words come before CHASM.Step 6: Now fix the first two letters as C and H. Remaining letters are A, S, M. The third letter in CHASM is A. There is no letter smaller than A among these, so add 0.Step 7: Fix CHA and remaining letters S and M. Sorted, they are M, S. The fourth letter in CHASM is S. Among M and S, the letter smaller than S is M, giving 1 additional word CHAM followed by the remaining letter.Step 8: Sum words before CHASM: 24 + 6 + 0 + 1 = 31. Therefore, CHASM is the 32nd word. Verification / Alternative check: Why Other Options Are Wrong: 24: This corresponds to the total number of words beginning with A, but ignores words starting with C A that precede CHASM. 36 and 72: These overestimate the rank and suggest incorrect counting, perhaps forgetting to sort letters correctly or miscomputing factorial contributions at each step. Common Pitfalls: Students often forget to sort the letters first before counting words starting with smaller first letters. Another frequent mistake is to ignore the influence of subsequent positions, or to use combinations instead of permutations. Careful tracking of remaining letters and using factorial counts at each position is crucial for accurate ranking. Final Answer: The rank of the word CHASM in dictionary order is 32.

Question 7

From the digits 1, 2, 3, 4, 5, 6 and 7, how many 6 digit even numbers can be formed such that no digit is repeated and the second last digit (tens place) is even?

Accepted Answer

Introduction / Context: This question is about constructing 6 digit even numbers from a given set of digits with no repetition. There is an additional constraint that the second last digit must be even. It tests understanding of positional restrictions and counting with permutations. Given Data / Assumptions: Available digits: 1, 2, 3, 4, 5, 6, 7. We must form 6 digit numbers. No digit may repeat in any number. The number must be even, so the units digit must be even. The second last digit (tens place) must also be even. Even digits available are 2, 4 and 6. Concept / Approach: We work position wise. The last digit must be even and the second last digit must also be even, and these two positions must have different even digits because repetition is not allowed. Once we choose and arrange these two even digits for the last two positions, we then arrange the remaining four positions (from the remaining five digits) in any order. Finally, we multiply the counts of these independent choices. Step-by-Step Solution: Step 1: Identify the even digits: 2, 4 and 6.Step 2: We must fill the last two places (tens and units) with 2 distinct even digits. This is a permutation of 3 even digits taken 2 at a time.Step 3: Number of ways to choose and arrange these two even digits for the last two positions is 3P2 = 3 * 2 = 6.Step 4: After using 2 of the 3 even digits, we have 5 digits left (since we started with 7 digits total).Step 5: The first four positions (lakhs, ten thousands, thousands, hundreds) must be filled with any 4 of the remaining 5 digits, with no repetition and no additional restrictions.Step 6: The number of ways to arrange 4 positions from 5 available digits is 5P4 = 5 * 4 * 3 * 2 = 120.Step 7: Combine the choices for the last two positions and the first four positions: total numbers = 6 * 120 = 720. Verification / Alternative check: Why Other Options Are Wrong: 521 and 225: These numbers do not arise from any natural product of permutations here and likely result from miscounting positions or mixing combinations with permutations. 420: This undercounts and might arise if someone uses 3C2 instead of 3P2 for the last two positions, ignoring the order of the two even digits. Common Pitfalls: Students often mistakenly use combinations instead of permutations for positional assignments, forgetting that the order of digits matters in a number. Some also forget that digits cannot repeat and allow the same even digit in both the second last and last position. Others may incorrectly constrain the leading digit, although here there is no ban on using 1 as the first digit. Being clear on each positional rule helps avoid these errors. Final Answer: The number of 6 digit even numbers that can be formed under the given conditions is 720.

Question 8

Eight distinct toys are to be distributed among 5 distinct children. If each toy can be given to any child and children may receive any number of toys (including zero), in how many different ways can this distribution be done?

Accepted Answer

Introduction / Context: This question is a distribution problem with distinct objects (toys) and distinct recipients (children). Each toy can go independently to any child, and there is no restriction on how many toys a child may get. It is a classic example of counting functions from one finite set to another. Given Data / Assumptions: There are 8 distinct toys. There are 5 distinct children. Each toy must be given to exactly one child. A child can receive any number of toys, including zero. Different assignments of toys to children count as different distributions. Concept / Approach: Think of each toy as making an independent choice of which child it goes to. For each toy there are 5 possible children. The choices for different toys are independent, so the total number of distributions is the product of the choices at each step. That naturally leads to a power expression of the form 5^8. Step-by-Step Solution: Step 1: Consider the first toy. It can be given to any of the 5 children, so there are 5 choices.Step 2: Consider the second toy. Regardless of who received the first toy, this toy again has 5 choices of children.Step 3: The same logic applies to each of the 8 toys. Each toy has 5 possible recipients.Step 4: Because the assignment of each toy is independent of the others, the total number of distributions is 5 * 5 * 5 * 5 * 5 * 5 * 5 * 5.Step 5: This product is 5^8. Verification / Alternative check: An alternative interpretation is to view each distribution as a function from the set of 8 toys to the set of 5 children, where each toy is mapped to exactly one child. The number of such functions from an 8 element set to a 5 element set is 5^8, which matches the counting argument above. You can also check with a small example, such as 2 toys and 3 children, where the correct answer is 3^2 = 9, and manually listing them will confirm the formula. Why Other Options Are Wrong: 5P8 and 8P5: These are permutation counts and would apply if we were arranging items in order, not assigning each toy to a child. 8^5: This would count functions in the opposite direction, from 5 children to 8 toys, which is not the situation here. Common Pitfalls: One common error is to incorrectly use combinations or permutations instead of recognising this as a simple independent choice problem. Another is to believe that each child must receive at least one toy, which would add an extra constraint and change the counting method completely. Carefully reading that children may receive any number of toys, including zero, is crucial to using the 5^8 formula. Final Answer: The number of ways to distribute the 8 distinct toys among 5 children is 5^8.

Question 9

In how many different ways can the letters of the word 'THERAPY' be arranged so that the two vowels always come together as a single block in every arrangement?

Accepted Answer

Introduction / Context: This problem tests permutations of letters of a word with a specific constraint on vowels. The word THERAPY consists of seven distinct letters, and the condition is that the vowels must always come together in every arrangement. Such questions are standard in permutation and combination topics and help you understand how to treat a group of items as a single block while still allowing internal rearrangements inside that block. Given Data / Assumptions: - Word: THERAPY. - Total letters: 7 (all distinct). - Vowels in the word: E and A (2 vowels). - Consonants: T, H, R, P, Y (5 consonants). - Vowels must always appear together, without any consonant in between them. - Order of letters matters (we are counting different permutations). Concept / Approach: The central idea is to treat the block of vowels as a single combined unit initially. If the vowels are forced to be together, we can think of the pair (E, A) as one object, along with the remaining consonants. After arranging this set of objects, we then count how many internal arrangements are possible for the vowels inside their block. The total number of valid arrangements is obtained by multiplying these two counts, because every arrangement of blocks can be paired with every internal arrangement of the vowels. Step-by-Step Solution: Step 1: Group the two vowels E and A together as a single unit, say [EA].Step 2: Now the units to arrange are: [EA], T, H, R, P, Y. This gives 6 distinct units in total.Step 3: The number of ways to arrange 6 distinct units in a row is 6!.Step 4: Compute 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.Step 5: Within the vowel block, the letters E and A can themselves be arranged in 2! ways (EA or AE).Step 6: Compute 2! = 2.Step 7: For each arrangement of the 6 units, there are 2 internal arrangements of the vowels, so total arrangements = 6! * 2! = 720 * 2 = 1440. Verification / Alternative check: An alternative way is to recognise that once vowels must stay together, you have effectively reduced the seven-letter problem to a six-letter one with an extra factor for internal vowel permutations. Since the letters are all distinct and there are no repeated letters, no further adjustments are needed. Recomputing 6! as 720 and multiplying again by 2 confirms the total as 1440. This agrees with the systematic method described above and fits the requirement that each valid arrangement has the vowels adjacent as one block. Why Other Options Are Wrong: - 720 counts only the permutations of six units and ignores the two internal arrangements of the vowels, undercounting by a factor of 2. - 1800 is larger than 1440 and does not correspond to any correct factorisation of the problem; it would require 6! * 2.5, which has no combinatorial meaning here. - 3600 is an overestimate and might come from multiplying by an incorrect extra factor, such as assuming more vowels or miscounting positions. Common Pitfalls: Students often forget that when a block is formed, its internal permutations must also be counted. Another frequent mistake is to assume the vowels are fixed in a single internal order, effectively taking only 1 arrangement instead of 2!. Additionally, some learners mistakenly treat the problem as unrestricted permutations of all seven letters (7! = 5040), failing to enforce the adjacency condition. Carefully separating the arrangement of blocks from internal arrangements prevents such errors. Final Answer: The letters of the word THERAPY can be arranged in 1440 different ways such that the vowels always come together.

Question 10

Find the number of distinct 4-element subsets that can be formed from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.

Accepted Answer

Introduction / Context: This problem is a straightforward application of combinations. We are asked to count how many different subsets of size 4 can be chosen from a fixed set of 11 distinct elements. Order inside a subset does not matter, which means we are not forming arrangements or permutations, but pure combinations. Such questions reinforce the idea of choosing items without regard to order, which is central to many counting problems in aptitude tests and mathematics competitions. Given Data / Assumptions: - Underlying set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. - Total number of elements in the set: 11. - We must form subsets that contain exactly 4 elements. - The order of elements in a subset does not matter. - Each element can either be included or excluded, but cannot be repeated. Concept / Approach: When we choose r elements from a set of n distinct elements without caring about order, we use the combination formula nCr. Here, n = 11 and r = 4. The formula is: nCr = n! / (r! * (n - r)!).We substitute n = 11 and r = 4 into this formula and simplify carefully. The result gives the total number of different 4-element subsets that can be formed. Step-by-Step Solution: Step 1: Identify n and r: n = 11, r = 4.Step 2: Use the formula 11C4 = 11! / (4! * 7!).Step 3: Rewrite 11! / 7! as 11 * 10 * 9 * 8, because 7! cancels from numerator and denominator.Step 4: So 11C4 = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1).Step 5: Compute the numerator: 11 * 10 = 110; 9 * 8 = 72; 110 * 72 = 7920.Step 6: Compute the denominator: 4 * 3 * 2 * 1 = 24.Step 7: Divide: 7920 / 24 = 330. Verification / Alternative check: You can check the result using a symmetry property: 11C4 is equal to 11C7, because choosing 4 elements to include is equivalent to choosing 7 elements to exclude. Many standard combination tables or calculators confirm that 11C4 = 330. Also, if you roughly estimate, 10C4 = 210 and 12C4 = 495, so a value between 210 and 495 is reasonable, and 330 fits nicely in this range. Why Other Options Are Wrong: - 340 and 370 are close but arise from arithmetic mistakes, such as mis-multiplying or mis-dividing when simplifying 11C4. - 320 is also a common slip when performing the division 7920 / 24 too quickly or truncating wrongly. Common Pitfalls: Common errors include confusing combinations with permutations and using 11P4 instead of 11C4, which would count ordered selections and give a much larger number. Another typical mistake is to expand the factorials incorrectly or not cancel common factors properly, leading to minor arithmetic errors. Writing out the product explicitly and cancelling factors step by step helps avoid such problems. Final Answer: The number of 4-element subsets that can be formed from the set {1, 2, ..., 11} is 330.

Question 11

How many arrangements of the letters of the word 'BENGALI' can be made if all the vowels appear only in odd positions of the arrangement (and never in even positions)?

Accepted Answer

Introduction / Context: This question combines permutations with position-based restrictions. The word BENGALI has seven distinct letters, with some vowels and some consonants. The restriction is that vowels must occupy only odd positions in the final arrangement, although odd positions may also contain consonants if some odd spots remain unused by vowels. This typical pattern appears frequently in placement and entrance exams to test understanding of counting under positional conditions. Given Data / Assumptions: - Word: BENGALI. - Total letters: 7, all distinct (B, E, N, G, A, L, I). - Vowels: E, A, I (3 vowels). - Consonants: B, N, G, L (4 consonants). - Positions available: 7, labeled 1 to 7. - Odd positions: 1, 3, 5, 7 (4 positions). - Even positions: 2, 4, 6 (3 positions). - Vowels must occupy only odd positions; they cannot appear in even positions. Concept / Approach: The vowels must be placed in odd positions only. There are 4 odd positions and 3 vowels, so we must choose 3 of the 4 odd positions for the vowels, then arrange the vowels among those chosen positions. After that, we place the 4 consonants in the remaining 4 positions (3 even positions plus the 1 unused odd position). Each stage is counted using combinations or permutations as appropriate, and the total is the product of these counts. Step-by-Step Solution: Step 1: Choose which 3 of the 4 odd positions will be occupied by vowels.Step 2: The number of ways to choose 3 positions out of 4 is 4C3 = 4.Step 3: Arrange the 3 vowels (E, A, I) in the chosen 3 positions. Since they are distinct, there are 3! = 6 ways.Step 4: Now 4 positions remain (the 3 even positions plus the 1 unused odd position), and we must place the 4 consonants (B, N, G, L) there.Step 5: The 4 consonants can be arranged in these 4 positions in 4! = 24 ways.Step 6: Total valid arrangements = 4C3 * 3! * 4! = 4 * 6 * 24.Step 7: Compute: 4 * 6 = 24, and 24 * 24 = 576. Verification / Alternative check: You can also reason by first placing vowels and then consonants without explicitly splitting between odd and even counts. The number of ways to assign vowels to odd positions and then fill the rest with consonants remains 4C3 * 3! * 4!, which yields 576. Trying a small example with fewer letters and positions can help confirm that this logic scales correctly to the full problem, increasing confidence in the computation. Why Other Options Are Wrong: - 720 is 6! or a related value and typically arises from forgetting the restriction on positions and counting all permutations or misusing factorials. - 567 is a near miss that might result from arithmetic slips rather than a correct combinatorial argument. - 625 does not factor naturally into products of small factorials and combinations for this problem and does not reflect any standard counting step here. Common Pitfalls: One frequent mistake is to assume that all odd positions must be occupied by vowels, which would be impossible here because there are 4 odd positions but only 3 vowels. Another error is to place vowels in any positions and then try to subtract disallowed cases, which makes the counting complicated and error-prone. Using the direct constructive approach—choose allowed positions for vowels and then permute consonants in the remaining slots—keeps the logic clear and avoids overcounting. Final Answer: The letters of the word BENGALI can be arranged in 576 different ways so that vowels appear only in odd positions.

Question 12

From a group of 5 consonants and 4 vowels, in how many different words can be formed using exactly 3 consonants and 2 vowels, assuming that order of letters in the word matters?

Accepted Answer

Introduction / Context: This question checks your understanding of combinations and permutations together. We are forming words (ordered arrangements of letters) from a specified pool of consonants and vowels, with the constraint that each selected word must contain exactly 3 consonants and 2 vowels. The problem explicitly involves choosing which letters to use and then arranging them, so both selection and ordering steps must be handled carefully. Given Data / Assumptions: - Available consonants: 5 distinct consonants. - Available vowels: 4 distinct vowels. - Each word must have exactly 5 letters in total. - Composition of each word: 3 consonants and 2 vowels. - No letter is repeated in a single word. - Order of letters in a word is important, so different permutations count as different words. Concept / Approach: The process naturally splits into two stages. First, choose which consonants and vowels will appear in the word. Second, arrange the chosen 5 letters in all possible orders. The number of ways to choose consonants uses combinations from the 5 available consonants, and likewise for vowels. After choosing 3 consonants and 2 vowels, we have 5 distinct letters, and these can be arranged in 5! different ways. The final count is the product of counts from these two main stages. Step-by-Step Solution: Step 1: Choose 3 consonants out of 5 available consonants.Step 2: Number of ways to do this is 5C3 = 10.Step 3: Choose 2 vowels out of 4 available vowels.Step 4: Number of ways to do this is 4C2 = 6.Step 5: After selections, we have 5 distinct letters (3 consonants + 2 vowels).Step 6: Arrange these 5 letters in all possible orders. The number of permutations is 5! = 120.Step 7: Multiply the three factors: 5C3 * 4C2 * 5! = 10 * 6 * 120.Step 8: Compute: 10 * 6 = 60; 60 * 120 = 7200. Verification / Alternative check: An alternate viewpoint is to select positions first for consonants or vowels and then fill them. However, because the letters themselves are distinct, this re-phrasing still leads to the same product 10 * 6 * 120. Using a small toy example with fewer letters helps demonstrate that each step corresponds to an independent choice, and the basic rule of product (multiplication principle) justifies multiplying the counts from each stage. Why Other Options Are Wrong: - 7600 and 6400 usually arise from arithmetic errors, such as miscomputing combinations or mis-evaluating 5! as 100 or 130 instead of 120. - 3600 might come from mistakenly dividing the total by 2 or halving at some step, perhaps wrongly assuming that order does not matter fully. Common Pitfalls: Some learners forget to separate the selection of letters from the arrangement and attempt to count everything directly as permutations, which can lead to confusion or double-counting. Others mistakenly treat combinations as if order mattered and use permutations in the selection phase. Maintaining a clear distinction—combinations for choosing which letters, permutations for ordering them—helps keep the logic straight and yields the correct result. Final Answer: The number of different words that can be formed using exactly 3 consonants and 2 vowels is 7200.

Question 13

A hotel has exactly three rooms: one single room (for 1 person), one double room (for 2 persons), and one four-person room (for 4 persons). In how many distinct ways can 7 different persons be housed in these rooms so that every bed is occupied?

Accepted Answer

Introduction / Context: This problem is about distributing distinct people into distinct rooms with fixed capacities. It is a typical application of combinations where each room has a fixed number of beds, and all people and rooms are distinguishable. The question asks for the number of ways to house seven persons in rooms of sizes 1, 2, and 4 such that each bed is filled. There is no mention of ordering within a room, so we consider only which persons share each room, not the seating arrangement inside the rooms. Given Data / Assumptions: - There are 7 distinct persons in total. - Rooms available: one single room (capacity 1), one double room (capacity 2), and one four-person room (capacity 4). - Total capacity: 1 + 2 + 4 = 7 beds, exactly matching the number of persons. - Rooms are distinct because they differ by type and capacity. - Order of people within the same room does not matter; only the grouping matters. Concept / Approach: The natural approach is to sequentially choose which person goes into the single room, which two go into the double room, and then allocate the remaining four to the four-person room. Each step uses combinations, as we are choosing subsets of people without regard to order inside the room. The multiplication principle applies because each independent choice can be combined with previous choices to form a complete allocation pattern. Step-by-Step Solution: Step 1: Choose 1 person out of 7 to occupy the single room.Step 2: The number of ways to choose this person is 7C1 = 7.Step 3: After assigning the single room, 6 persons remain for the other rooms.Step 4: Choose 2 of the remaining 6 persons to occupy the double room.Step 5: The number of ways to do this is 6C2 = 15.Step 6: The remaining 4 persons automatically go into the four-person room; there is only 1 way to choose them once the first two selections are fixed.Step 7: Total number of ways = 7C1 * 6C2 * 1 = 7 * 15 * 1 = 105. Verification / Alternative check: We can confirm by checking that no other hidden arrangements are being ignored. Since we do not distinguish positions within a room (for example, who sleeps in which specific bed of the four-person room), we correctly use combinations. If beds within rooms were labelled, we would need permutations inside each room, which would produce a larger result. Here, the room labels alone distinguish the groups, so 105 is the correct and complete count. Why Other Options Are Wrong: - 7!/5! equals 7 * 6 = 42 and would correspond to picking only two ordered positions, not full room assignments. - 420 is 4 times the correct answer and might come from an incorrect permutation treatment of the double room or four-person room. - 7! x 6! is astronomically too large and reflects a misunderstanding of the structure of the problem, effectively permuting people multiple times unnecessarily. Common Pitfalls: Students sometimes mistakenly treat ordering within rooms as significant, multiplying by extra factorials, or forget that rooms themselves are already distinguished by size. Another error is to choose the group of four first and then overcount by considering the same grouping multiple times in different orders. Sticking to a clean sequence—single room, then double room, then the rest—ensures each allocation is counted exactly once. Final Answer: The seven persons can be housed in 105 distinct ways in the three rooms.

Question 14

In how many different ways can the letters of the word 'OLIVER' be arranged so that the vowels in the word always appear in alphabetical (dictionary) order from left to right?

Accepted Answer

Introduction / Context: This problem focuses on permutations of letters with a restriction on the relative order of vowels. The word OLIVER contains six distinct letters, including three vowels. The condition is that whenever the vowels appear from left to right in any arrangement, they must follow dictionary order, meaning E must come before I, and I must come before O. Their positions in the word may vary, but their left-to-right order must always be E, then I, then O. This type of constraint is common in permutation questions to test understanding of relative ordering. Given Data / Assumptions: - Word: OLIVER. - Total letters: 6 distinct letters (O, L, I, V, E, R). - Vowels: O, I, E (3 vowels). - Consonants: L, V, R (3 consonants). - We must count permutations where, reading from left to right, the vowels appear in alphabetical order E, I, O, although other letters can appear in between. Concept / Approach: Without any restriction, all 6 letters can be arranged in 6! different ways. In those unrestricted permutations, the 3 vowels can appear in any of 3! possible relative orders with equal frequency. However, only 1 of these orders (E, I, O) satisfies the dictionary order condition. Therefore, the number of valid permutations is the total permutations divided by the number of possible vowel orders, effectively 6! / 3!. This is a powerful symmetry argument that avoids direct position-by-position counting. Step-by-Step Solution: Step 1: Compute the total number of permutations of 6 distinct letters with no restriction.Step 2: Total permutations = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.Step 3: Among these permutations, the vowels E, I, O can appear in 3! = 6 different relative orders: EIO, EOI, IEO, IOE, OEI, OIE.Step 4: Each of these 6 vowel-order patterns appears in exactly the same number of permutations, because all letters are distinct and symmetric in the counting.Step 5: We want only the permutations where vowels appear in the single order E, I, O.Step 6: Therefore, valid permutations = total permutations / number of vowel orders = 6! / 3! = 720 / 6.Step 7: Compute 720 / 6 = 120. Verification / Alternative check: An alternative method is to choose positions for the three vowels and then place the consonants, explicitly ensuring that the vowel positions follow E, I, O. First choose any 3 positions out of 6 for the vowels, which can be done in 6C3 ways. For each such choice, the vowels must occupy these positions in the fixed order E, I, O, so there is only 1 way to place vowels. The remaining 3 positions are filled with 3 consonants in 3! ways. Thus, total arrangements = 6C3 * 3! = 20 * 6 = 120, matching the earlier argument. Why Other Options Are Wrong: - 144 and 136 are close but arise from incorrect combinatorial factors or incomplete use of factorials. - 186 is not consistent with any natural factorisation like 6C3 * 3! or 6!/3! for this situation and therefore does not match the proper counting logic. Common Pitfalls: Students sometimes enforce that the vowels must be adjacent or occupy specific positions, which is not required here; only their relative order matters. Others wrongly count only one vowel ordering and forget to divide the total by 3!, or they attempt to overcomplicate the positional selection. Using symmetry and the idea of equally likely vowel orderings is the cleanest and most reliable approach. Final Answer: The letters of the word OLIVER can be arranged in 120 different ways so that the vowels appear in dictionary order from left to right.

Question 15

Nine different letters of the alphabet are given and five-letter words are formed using these letters. How many such five-letter words can be formed which contain at least one repeated letter?

Accepted Answer

Introduction / Context: This question is about counting words (ordered strings of letters) allowing repetition and then restricting to those that actually have at least one repeated letter. Problems of this type are best handled using the complement method: first count all possible words, then subtract those with no repeated letters. This is a classic application of permutations with repetition and the principle of complementary counting in combinatorics. Given Data / Assumptions: - There are 9 distinct letters available. - We are forming five-letter words (strings) using these letters. - Repetition of letters is allowed in general, but we want only those words with at least one letter repeated. - Order matters: two strings that differ in position of letters are considered different words. Concept / Approach: Directly counting all words that have at least one repeated letter is complicated, because many overlapping cases would need to be included. Instead, we use the complement approach. We first count all possible five-letter words that can be formed using the 9 letters with repetition allowed, and then subtract the number of words with no repeated letters at all (that is, all letters distinct). The result is the count of words where at least one letter appears more than once. Step-by-Step Solution: Step 1: Count the total number of five-letter words that can be formed from 9 letters when repetition is allowed.Step 2: For each of the 5 positions in the word, there are 9 choices (any of the 9 letters).Step 3: Total unrestricted words = 9^5 = 9 * 9 * 9 * 9 * 9 = 59049.Step 4: Next, count the number of five-letter words in which all letters are distinct (no repetition).Step 5: This is a permutation without repetition: 9P5 = 9 * 8 * 7 * 6 * 5.Step 6: Compute 9P5: 9 * 8 = 72; 72 * 7 = 504; 504 * 6 = 3024; 3024 * 5 = 15120.Step 7: Words with at least one repeated letter = total words - words with all distinct letters.Step 8: So required count = 59049 - 15120 = 43929. Verification / Alternative check: We can approximate the proportion of words with repeated letters by noting that 15120 is much smaller than 59049, so most words will indeed contain repetitions. Computing the difference gives 43929, and a quick check of the arithmetic 59049 - 15000 = 44049, then subtracting another 120 gives 43929, confirming the subtraction. This aligns with the logical expectation that only a small fraction of all possible five-letter words will have all distinct letters when repetition is allowed. Why Other Options Are Wrong: - 59049 counts all possible words, including those with no repetition; it ignores the condition of having at least one repeated letter. - 15120 counts only words with all distinct letters, that is, the complement of what is required. - 0 would suggest that no word has repetition, which is obviously incorrect when repetition is allowed in 9^5 possibilities. Common Pitfalls: A common mistake is to misinterpret the phrase "at least one letter repeated" and try to count cases of exactly one repetition, two repetitions, and so on separately, which becomes extremely complicated and prone to double-counting. The complement method efficiently avoids this complexity. Another error is confusing 9P5 with 9^5, or assuming that 9P5 already counts repeated letters, which it does not. Being clear on when repetition is allowed versus disallowed is crucial. Final Answer: The number of five-letter words that can be formed from 9 distinct letters with at least one repeated letter is 43929.

Question 16

To fill 8 vacancies there are 15 candidates, of whom 5 belong to the ST category. If 3 of the vacancies are reserved for ST candidates and the remaining 5 vacancies are open to all candidates, in how many different ways can the selections be made?

Accepted Answer

Introduction / Context: This question involves selection with a reservation constraint. There are 15 candidates, including 5 ST candidates, and 8 vacancies to fill. Exactly 3 vacancies are reserved for ST candidates, and the remaining 5 are open to all, which effectively means they can be filled by any of the remaining candidates, regardless of category. This structure appears frequently in competitive exams where reserved and open positions are considered within the same selection process. Given Data / Assumptions: - Total candidates: 15. - ST candidates: 5. - Non-ST candidates: 10 (since 15 - 5 = 10). - Total vacancies: 8. - 3 vacancies are reserved and must be filled by ST candidates. - The remaining 5 vacancies are open to all remaining candidates (both ST and non-ST). - We only care about which candidates are selected, not the specific seat they occupy. Concept / Approach: The selection can be naturally decomposed into two stages. First, we choose the 3 ST candidates to fill the reserved seats. Next, we fill the remaining 5 open seats from all remaining candidates (which include the remaining ST candidates and all non-ST candidates). Because seats themselves are not distinguished beyond the reserved versus open category, we only use combinations for each choice and multiply the results according to the rule of product. Step-by-Step Solution: Step 1: Select 3 ST candidates out of the 5 available ST candidates to fill the reserved vacancies.Step 2: Number of ways to do this is 5C3 = 10.Step 3: After choosing these 3 ST candidates, the remaining candidate pool consists of 15 - 3 = 12 people (2 ST and 10 non-ST).Step 4: We now must select 5 additional candidates from this pool of 12 to fill the open vacancies.Step 5: The number of ways to select 5 out of 12 candidates is 12C5.Step 6: Compute 12C5 using the formula 12! / (5! * 7!). This simplifies to (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1) = 792.Step 7: Total ways to complete the selection = 5C3 * 12C5 = 10 * 792 = 7920. Verification / Alternative check: You can verify the logic by considering that the reserved seats force exactly 3 ST candidates to be selected initially. After that, the open seats simply require choosing any 5 from the 12 remaining candidates. No double-counting occurs because once a candidate is selected for the reserved group, they are not available for the second selection. Recomputing 12C5 independently or using known values from combination tables confirms it is 792, giving 7920 total ways when multiplied by 10. Why Other Options Are Wrong: - 74841 and 14874 do not arise from any natural combination product for this structure; they reflect miscalculations of combinations or incorrect interpretations of the reservation rule. - 10213 is not a factorable product of small combination numbers like 5C3 and 12C5 and therefore does not correspond to a correct counting argument here. Common Pitfalls: One common error is to treat all 8 vacancies as open and then try to enforce the ST requirement afterward, risking overcounting and inclusion of selections with fewer than 3 ST candidates. Another mistake is to assume that open vacancies cannot be filled by ST candidates, which is not stated in the question. Carefully reading that open seats are available to all candidates ensures that the remaining ST candidates are still part of the pool for the open positions. Final Answer: The number of ways to select candidates is 7920.

Question 17

Seven teachers and six students are to be seated around a round table. In how many different circular arrangements can they sit so that no two students are seated next to each other?

Accepted Answer

Introduction / Context: This is a circular permutation problem with restrictions. We have 7 teachers and 6 students, all distinct individuals, to be seated around a round table. The key condition is that no two students may sit next to each other. Circular arrangements are counted up to rotation (that is, rotating everyone together is considered the same arrangement), and this changes the counting formula compared to linear arrangements. The “gap method” is the standard approach for problems where a subset of people cannot sit together. Given Data / Assumptions: - Distinct people: 7 teachers (T) and 6 students (S), total 13 persons. - Seating is around a circular table, so rotations of the same relative order are not counted separately. - No two students may sit adjacent to each other. - All teachers and students are distinguishable from one another. Concept / Approach: To ensure that no two students are adjacent, we first arrange the teachers around the table and then place the students in the “gaps” between teachers. Using circular permutation logic, we fix the relative order of teachers first. Once teachers are seated, they create specific gaps around the table where students can be placed without violating the adjacency rule. Each gap can contain at most one student if we want to avoid two students sitting together. Step-by-Step Solution: Step 1: Arrange the 7 teachers around the circular table.Step 2: The number of distinct circular arrangements for 7 distinct teachers is (7 - 1)! = 6!.Step 3: Once teachers are seated, there are exactly 7 gaps between them (one between each pair of adjacent teachers, going around the circle).Step 4: To prevent students from sitting together, at most one student may occupy each gap.Step 5: We must place 6 distinct students into these 7 gaps, choosing 6 of the 7 gaps and ordering the students among those chosen gaps.Step 6: The number of ways to choose which 6 gaps will be occupied is 7C6 = 7.Step 7: Once the 6 gaps are chosen, the 6 students can be permuted among these 6 positions in 6! ways.Step 8: Total number of valid circular arrangements = (arrangements of teachers) * (ways to select gaps) * (ways to permute students).Step 9: Thus, total arrangements = 6! * 7C6 * 6! = 6! * 7 * 6!.Step 10: Recognise that 7! = 7 * 6!, so 6! * 7 * 6! = 7! * 6!. Verification / Alternative check: We can sanity-check the formula by observing that there are enough teacher “separators” for the students. There are 7 gaps and only 6 students, so it is indeed possible to seat all students without adjacency. If we had 7 students instead, it would be impossible, as we would need to fill every gap, forcing two students to become adjacent when the circle closes. The final expression 7! * 6! exactly matches option 7! x 6!, confirming the correct combination of gap selection and student permutation. Why Other Options Are Wrong: - 7! x 7! suggests overcounting, effectively permuting students into 7 positions instead of 6, which is not needed and violates the constraint of having only 6 students. - 6! x 6! ignores the 7 choices of which gap remains empty and thus undercounts the possible configurations. - 7! x 5! similarly fails to respect the number of student permutations and gap selection structure, providing an incorrect factor. Common Pitfalls: Common mistakes include treating the arrangement as linear rather than circular, leading to an extra factor of 13 from rotations; or placing students first and then trying to insert teachers, which complicates the condition and often causes overcounting or invalid configurations. Forgetting that each gap can hold at most one student is another error that leads to arrangements where students are adjacent. Using the teachers as fixed separators and then assigning students to gaps systematically avoids these issues. Final Answer: The number of circular arrangements in which no two students sit together is 7! × 6!.

Question 18

In a box there are 5 black pens, 3 white pens and 4 red pens. In how many different ways can we choose exactly 2 black pens, 2 white pens and 2 red pens from this box?

Accepted Answer

Introduction / Context: This is a direct combinations problem involving choosing items of different colours with exact counts from each colour group. The pens of the same colour are considered identical for selection purposes only in terms of colour count; however, since they are simply “pens”, we treat each pen as distinct but constrained by colour. The key idea is to use the combination formula separately for each colour group and then multiply the results because each selection of black pens can be combined with each selection of white and red pens independently. Given Data / Assumptions: - Number of black pens: 5. - Number of white pens: 3. - Number of red pens: 4. - Required selection: exactly 2 black pens, 2 white pens, and 2 red pens. - The order in which we pick pens does not matter; only which pens are in the final group matters. Concept / Approach: We handle each colour separately using combinations. From each colour group, we must choose a fixed number of pens. Once we compute the number of ways to choose each colour group, we multiply those counts together by the rule of product because the choices for black, white and red pens are independent of one another. No permutations are needed because we are not arranging the chosen pens in any sequence; we only form a set of six pens obeying the colour constraint. Step-by-Step Solution: Step 1: Choose 2 black pens from 5 black pens. The number of ways is 5C2.Step 2: Compute 5C2 = (5 * 4) / (2 * 1) = 10.Step 3: Choose 2 white pens from 3 white pens. The number of ways is 3C2.Step 4: Compute 3C2 = (3 * 2) / (2 * 1) = 3.Step 5: Choose 2 red pens from 4 red pens. The number of ways is 4C2.Step 6: Compute 4C2 = (4 * 3) / (2 * 1) = 6.Step 7: Total number of valid selections = 5C2 * 3C2 * 4C2 = 10 * 3 * 6.Step 8: Compute 10 * 3 = 30, and then 30 * 6 = 180. Verification / Alternative check: We can validate by reasoning that any choice of 2 black, 2 white and 2 red pens is uniquely determined by which specific pens were chosen from each colour group. There is no overlap between colours, so there is no double-counting when multiplying the separate combination counts. Re-calculating quickly, 5C2 = 10, 3C2 = 3, 4C2 = 6, and their product 10 * 3 * 6 is clearly 180, which matches the computed result. Why Other Options Are Wrong: - 220 and 240 are higher than the correct total and likely originate from miscomputing one or more combinations or accidentally including arrangements where the colour counts are not exactly 2 each. - 160 is smaller than the correct answer and might result from omitting the red or white selection factor or using 4C2 = 4 instead of 6. Common Pitfalls: Some students mistakenly use permutations instead of combinations, multiplying by additional factorials, which would correspond to ordering the selected pens. Others may count only one colour correctly and miscalculate the others. A third common mistake is to try to treat all pens as identical except by colour and then overthink the structure. The straightforward approach is: use combinations colour by colour, then multiply. Final Answer: There are 180 different ways to choose exactly 2 black, 2 white and 2 red pens from the box.

Question 19

How many six-digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8 and 9 if no digit is repeated in any number?

Accepted Answer

Introduction / Context: This question asks for the count of six-digit odd numbers formed from a specific set of digits where repetition is not allowed. It tests understanding of both place value restrictions (odd number means last digit must be odd) and the basic permutation counting rules with exclusions on leading zeros. Paying attention to which digits can occupy the units and thousands places is essential. Given Data / Assumptions: - Available digits: 0, 2, 3, 5, 6, 7, 8, 9 (8 distinct digits in total). - We must form six-digit numbers (first digit cannot be zero). - Each digit can appear at most once (no repetition allowed). - The number must be odd, so the last digit must be one of the available odd digits. - Odd digits available: 3, 5, 7, 9 (4 digits). Concept / Approach: We first determine the possible choices for the last digit because of the odd-number condition. After fixing the last digit, we then choose the first digit, remembering that it cannot be zero. The remaining four middle positions can then be filled with any of the remaining digits without repetition. We proceed step by step, counting the number of options at each position and multiplying using the rule of product. Step-by-Step Solution: Step 1: Choose the units (last) digit. Since the number must be odd, the last digit must be 3, 5, 7, or 9.Step 2: There are 4 choices for the last digit.Step 3: Once the last digit is chosen, it cannot be used again. There remain 7 unused digits, including zero.Step 4: Choose the first (leftmost) digit from these 7 remaining digits, but it cannot be zero because we need a six-digit number.Step 5: Out of the 7 remaining digits, exactly 1 is zero, so there are 7 - 1 = 6 valid choices for the first digit.Step 6: After fixing the first and last digit, there are 6 unused digits left to fill the 4 middle positions.Step 7: The middle four positions can be filled by these 6 digits without repetition in 6P4 ways (permutations of 6 taken 4 at a time).Step 8: Compute 6P4 = 6 * 5 * 4 * 3 = 360.Step 9: Total number of valid numbers = (choices for last digit) * (choices for first digit) * (arrangements of middle digits) = 4 * 6 * 360.Step 10: Compute 4 * 6 = 24; 24 * 360 = 8640. Verification / Alternative check: We can confirm the reasoning by considering the total six-digit numbers that can be formed without the oddness restriction and then restricting the last digit to the 4 odd ones. The step-by-step approach above ensures we never assign zero to the first place and never reuse digits. The final multiplication 4 * 6 * 360 is straightforward, and recomputing 6P4 as 6 * 5 * 4 * 3 = 360 confirms no arithmetic mistake has been made. Why Other Options Are Wrong: - 720 is far too small and might come from counting only 6! permutations, ignoring the restrictions. - 3620 and 4512 are intermediate-like numbers but do not match any natural product of 4, 6 and a permutation term; they likely reflect mis-handling of the zero constraint or incomplete multiplication. Common Pitfalls: Students often forget that the first digit of a six-digit number cannot be zero, leading them to count some invalid numbers. Another error is to treat the middle four positions as combinations instead of permutations, using 6C4 instead of 6P4, which ignores the importance of order in forming different numbers. Ensuring that each positional choice respects both non-repetition and the number structure avoids such mistakes. Final Answer: There are 8640 six-digit odd numbers that can be formed from the given digits without repetition.

Question 20

On the occasion of New Year, each of the 21 students in a class sends a greeting card to every other student in the class. What is the total number of greeting cards exchanged among the students?

Accepted Answer

Introduction / Context: This is a classic counting problem involving pairwise exchanges between students. Each student sends greeting cards to all others, so we must count the total number of directed cards (from sender to receiver). It tests the understanding of how many ordered pairs of distinct students can be formed from a given group size. This kind of thinking appears in graph theory (directed edges), handshake problems, and simple combinatorial reasoning. Given Data / Assumptions: - Total number of students in the class: 21. - Each student sends a greeting card to every other student. - No student sends a card to themselves. - Cards are directed: a card from student A to student B is distinct from a card from B to A. Concept / Approach: For each student, count how many cards they send, then sum over all students. Since every student has the same number of recipients, we can multiply. Each student sends a card to every other student, which means to 21 - 1 = 20 other students. There are 21 students in total, so the total number of cards is 21 multiplied by 20. This directly uses the multiplication principle: (number of possible senders) * (number of possible receivers for each sender). Step-by-Step Solution: Step 1: Fix one student as the sender.Step 2: That student sends greeting cards to all remaining 20 students in the class.Step 3: So each student sends 20 cards.Step 4: There are 21 students acting as senders.Step 5: Multiply to get the total number of cards: 21 * 20.Step 6: Compute 21 * 20 = 420. Verification / Alternative check: We can also think in terms of ordered pairs (sender, receiver). The number of ordered pairs of distinct students from a set of 21 is 21 * 20, because there are 21 choices for the sender and then 20 remaining choices for the receiver. This is exactly the same calculation as above. If the problem had asked for undirected exchanges (counting a pair of students only once, regardless of who sends to whom), we would divide by 2, but here each direction is a separate card, so we keep 21 * 20. Why Other Options Are Wrong: - 380 and 400 are less than 420 and could arise from subtracting some pairs or miscounting the number of recipients per student. - 441 is 21^2 and would incorrectly include pairs where a student sends a card to themselves, which is not allowed in this scenario. Common Pitfalls: Some learners mix up directed and undirected counting. If one thinks only in terms of pairs of students exchanging cards, they might compute 21C2, which gives 210, and then double it or forget to double it. It is important to interpret that “each student sends a card to every other student” means we care about direction: A to B and B to A are two separate cards. Keeping the distinction between ordered and unordered pairs in mind avoids confusion. Final Answer: The total number of greeting cards exchanged among the 21 students is 420.

Permutation and Combination Questions