Question 1

From the digits 1, 0, 2, 3, 5 and 6, how many different five digit numbers can be formed that are between 50000 and 60000, if no digit is repeated within a number?

Accepted Answer

Introduction / Context: This problem is a digit arrangement question with a range restriction. We must form five digit numbers from a given set of digits, obeying two key conditions: the number falls within a specific interval and digits are not repeated. Such questions test whether students can correctly apply constraints on leading digits while still using permutations for the remaining positions. Given Data / Assumptions: Available digits: 1, 0, 2, 3, 5 and 6. We must form five digit numbers. Digits cannot be repeated within a number. The number must be between 50000 and 60000, inclusive of 50000 and exclusive of 60000. Concept / Approach: A five digit number lies between 50000 and 60000 if and only if its first digit (ten thousands place) is 5. Thus, the leading digit is fixed. After fixing the first digit, we simply need to permute the remaining digits in the remaining positions, taking care not to repeat any digit. Since zero is allowed in positions other than the first, there is no restriction on placing 0 among the last four digits. Step-by-Step Solution: Step 1: Fix the first digit as 5, because only numbers starting with 5 are between 50000 and 60000 when using these digits.Step 2: After using 5, the remaining digits available for the last four places are {0, 1, 2, 3, 6}, which are five distinct digits.Step 3: We need to form all possible four digit sequences from these 5 digits without repetition for the remaining positions.Step 4: The number of ways to choose and arrange 4 positions from 5 distinct digits is 5P4 = 5 × 4 × 3 × 2.Step 5: Compute 5 × 4 × 3 × 2 = 120.Step 6: Each such arrangement corresponds to a unique five digit number starting with 5 and hence lying between 50000 and 60000. Verification / Alternative check: We can check that there is no additional condition affecting the last four digits. Since zero is not allowed as the first digit but is freely permitted in other positions, our count is exhaustive. Also, if we tried an alternative count by first choosing which digit is left unused and then arranging the others, we would get the same total: 5 ways to choose the omitted digit times 4! ways to arrange the remaining digits, giving 5 × 24 = 120. Why Other Options Are Wrong: Values like 240 or 360 would arise if we mistakenly allowed repetition of digits or misapplied permutation formulas. 60 would correspond to halving the correct count, which might happen if one incorrectly restricts the positions of zero. Only 120 correctly reflects the number of valid five digit numbers satisfying all conditions. Common Pitfalls: A common mistake is to forget that the first digit is forced to be 5 by the given range, and instead treat all digits as freely permutable. Another error is misplacing restrictions on zero and assuming it cannot appear at all, which would unnecessarily reduce the count. Careful consideration of each place value and the stated range prevents these errors. Final Answer: The number of five digit numbers between 50000 and 60000 that can be formed is 120.

Question 2

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can the selection be made so that at least one boy is included among the four selected children?

Accepted Answer

Introduction / Context: This is a classic selection problem involving a group of boys and girls with a condition that prohibits selecting all girls. It focuses on using combinations and the complement method, which is often simpler than direct counting of all allowed cases. Such questions are very common in the combinations part of aptitude tests. Given Data / Assumptions: There are 6 boys. There are 4 girls. Total children available: 10. We need to select 4 children. The selection must contain at least one boy, so an all girl selection is not allowed. Concept / Approach: Instead of counting all valid combinations directly by splitting into cases based on how many boys are chosen, it is simpler to count all possible selections of 4 children from the 10 and then subtract the number of invalid selections, i.e., those that contain no boys (all girls). This is a straightforward application of the complement principle in counting. Step-by-Step Solution: Step 1: Total ways to choose any 4 children from 10 is 10C4.Step 2: Compute 10C4 = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 210.Step 3: To violate the condition, a selection would have to consist of girls only, that is 4 girls with no boys.Step 4: The number of ways to choose 4 girls from the 4 available girls is 4C4 = 1.Step 5: Therefore, the number of valid selections with at least one boy is 210 - 1 = 209.Step 6: Hence, there are 209 different ways to select 4 children with at least one boy. Verification / Alternative check: We can verify this briefly by counting in cases: exactly 1 boy, 2 boys, 3 boys, and 4 boys. The total would be 6C1 × 4C3 + 6C2 × 4C2 + 6C3 × 4C1 + 6C4 × 4C0. Computing these gives 6 × 4 + 15 × 6 + 20 × 4 + 15 × 1 = 24 + 90 + 80 + 15 = 209. This matches our complement method result and confirms correctness. Why Other Options Are Wrong: 210 is the total number of ways without the restriction and thus includes the one invalid all girl selection. 200, 208 and 290 do not match either the complement calculation or the detailed case based counting and therefore are incorrect. Only 209 satisfies the constraint and matches both methods of calculation. Common Pitfalls: Students sometimes forget to exclude the all girl case or miscalculate combinations, especially 10C4. Another common mistake is to misinterpret the phrase "at least one boy" and treat it as "exactly one boy," which yields a much smaller number. Always read the condition carefully and choose the simplest counting strategy, which here is the complement method. Final Answer: The number of ways to select 4 children with at least one boy is 209.

Question 3

How many different arrangements can be made out of the letters of the word DRAUGHT if the two vowels in the word are never separated, that is, the vowels always occur together as a single block in every arrangement?

Accepted Answer

Introduction / Context: This is a permutations question involving a restriction on how vowels can appear. We must arrange the letters of a word under the condition that the vowels are always together. Such problems are common in competitive exams and are a direct application of treating a group of letters as a single block to simplify the arrangement count. Given Data / Assumptions: The word is DRAUGHT, containing the letters D, R, A, U, G, H and T. Total letters: 7, all distinct. The vowels in the word are A and U. The condition is that the vowels must never be separated; they must appear together in every arrangement. Concept / Approach: When vowels must stay together, we treat the group of vowels as one combined block. This reduces the problem to arranging that block along with the remaining consonants. After arranging the blocks, we then count the internal arrangements of the letters within the vowel block. The total number of valid arrangements is the product of these two counts. Step-by-Step Solution: Step 1: Identify vowels: A and U are vowels; the remaining letters D, R, G, H and T are consonants.Step 2: Treat the pair of vowels AU as one block. Along with the five consonants D, R, G, H and T, we have 6 objects to arrange: [AU], D, R, G, H, T.Step 3: The number of ways to arrange 6 distinct objects is 6!.Step 4: Compute 6! = 720.Step 5: Inside the vowel block, A and U can be arranged in 2! ways: AU or UA.Step 6: Total arrangements where the vowels stay together = 6! × 2! = 720 × 2 = 1440. Verification / Alternative check: If there were no restriction, the total number of arrangements of 7 distinct letters would be 7! = 5040. We have restricted our arrangements to those where the two vowels are together, giving 1440 possibilities. This is a subset of the full 5040 and seems reasonable in size. A manual check on a smaller analogous example supports the logic of treating the vowels as one block. Why Other Options Are Wrong: 720 would correspond to the case where we treat the vowel block as a single object but do not consider that the vowels could be internally permuted. 360 and 640 arise from partial or incorrect application of factorials. Only 1440 correctly multiplies the arrangements of the blocks by the internal arrangements of the vowels. Common Pitfalls: Students often forget to multiply by the number of internal arrangements of the vowels after treating them as a block. Another mistake is misidentifying the vowels or miscounting the total number of letters. Carefully listing letters and following the block method step by step avoids these errors. Final Answer: The number of arrangements where the vowels in DRAUGHT are never separated is 1440.

Question 4

A team of 8 students is going on an excursion in two cars. One car has a seating capacity of 5 and the other car has a seating capacity of 4. Since there are only 8 students and 9 seats, one seat will remain empty. In how many different ways can the…

Accepted Answer

Introduction / Context: This question tests the understanding of distributing people into distinct groups with capacity constraints. We must assign 8 students to two cars of capacities 5 and 4, with one seat necessarily left vacant, and count the total number of valid distributions. This is a combinatorial distribution problem frequently seen in reasoning and aptitude tests. Given Data / Assumptions: There are 8 distinct students. Car A has 5 seats and Car B has 4 seats. Total seats available: 9. Since there are only 8 students, exactly one seat will be empty. We are counting distinct ways of assigning students to cars, ignoring the specific seat positions within each car. Concept / Approach: We consider how many students go in each car under the seating constraints. There are two possible occupancy patterns consistent with 8 students and car capacities 5 and 4: Car A has 5 students and Car B has 3 students, or Car A has 4 students and Car B has 4 students. For each case, we count how many subsets of students fit the described distribution and sum the counts because the cases are mutually exclusive. Step-by-Step Solution: Step 1: Case 1: Car A carries 5 students, Car B carries 3 students (one vacant seat in Car B). Choose 5 students out of 8 for Car A.Step 2: The number of ways to choose 5 students for Car A is 8C5 = 56. The remaining 3 automatically go to Car B.Step 3: Case 2: Car A carries 4 students, Car B carries 4 students (one vacant seat in Car A). Choose 4 students out of 8 for Car B or Car A, the count is symmetric.Step 4: The number of ways to choose 4 students for Car B is 8C4 = 70. The remaining 4 go to Car A.Step 5: The two cases are disjoint, so total assignments = 56 + 70 = 126.Step 6: Therefore, there are 126 different ways for the students to travel in the two cars. Verification / Alternative check: We can also think in terms of choosing which seat remains empty, but since we are not distinguishing individual seats within the cars, this would overcomplicate the count. The combinational approach based on how many students ride in each car is simpler and directly uses the capacities to find all possibilities. The count 126 lies between 2^8 and 3^8 and seems reasonable for two groupings with capacity constraints. Why Other Options Are Wrong: 120 is very close but would correspond to a different counting structure, such as ignoring one of the valid occupancy patterns. Values like 146 or 156 do not arise from any natural combination of the two case counts. Only 126 exactly equals 8C5 + 8C4 and correctly accounts for both possible distributions of students between the two cars. Common Pitfalls: Students sometimes assume that Car A must always be full, ignoring the possibility that Car B could be full instead. Another common error is to consider 8C5 only, leading to 56 and missing the second case entirely. Recognising that one seat can be empty in either car helps identify both valid patterns. Final Answer: The number of different ways in which the 8 students can travel in the two cars is 126.

Question 5

From a group of six men and four ladies (a total of ten people), a committee of three members is to be formed. Mrs X is not willing to join any committee in which Mr Y is a member, whereas Mr Y is willing to join a committee only if Mrs Z is also in…

Accepted Answer

Introduction / Context: This is a conditional combinations problem involving three specific people with constraints on which of them can serve together on a committee. It tests the ability to interpret verbal restrictions correctly and to translate them into mathematical conditions. Such questions are common in reasoning sections as they combine logical conditions with combinatorial counting. Given Data / Assumptions: Total people: 6 men and 4 ladies, for a total of 10 individuals. Special persons: Mrs X, Mrs Z and Mr Y are among these ten. A committee of 3 members is to be formed. Mrs X will not join any committee that includes Mr Y. Mr Y will join a committee only if Mrs Z is also on that committee. We assume that Mrs X and Mrs Z are different ladies, and Mr Y is one of the men. Concept / Approach: We break the counting into cases based on whether Mr Y is in the committee or not. When Y is included, both conditions involving X and Z must be enforced. When Y is excluded, only the first condition becomes irrelevant and the committee selection is unrestricted. We then add the counts from the disjoint cases to get the total number of valid committees. Step-by-Step Solution: Step 1: Case 1: Committees that include Mr Y. Since Y is in the committee, Mrs Z must also be in the committee, and Mrs X cannot be included.Step 2: When Y is in, Z must be in, and X must be out. So two members are fixed: Y and Z. We need one more member from the remaining people (excluding X, Y and Z).Step 3: There are 10 total people minus these 3, leaving 7 people who can fill the third spot. So the number of committees in Case 1 is 7C1 = 7.Step 4: Case 2: Committees that do not include Mr Y. Then the conditions about X and Z involving Y do not apply, and we can choose any 3 people from the remaining 9 individuals (since Y is excluded).Step 5: Number of such committees is 9C3 = (9 × 8 × 7) / (3 × 2 × 1) = 84.Step 6: Total valid committees = committees with Y + committees without Y = 7 + 84 = 91. Verification / Alternative check: We can run a mental check: there are 10C3 = 120 possible committees without restrictions. We exclude those that violate the conditions. The derived count 91 means 29 committees are invalid. Quick reasoning shows that invalid committees are exactly those containing Y without Z or those containing both X and Y together. Counting these invalid combinations separately also leads to 29 and confirms the total of 91 valid committees. Why Other Options Are Wrong: Values like 98, 104, or 109 arise from miscounting one of the cases or forgetting the exclusion of X when Y is present. Only 91 is consistent with a careful case based analysis and with subtracting invalid committees from the unrestricted total of 120. Common Pitfalls: Students often misinterpret the statement about Y joining only if Z is included and may treat it as a biconditional (Z must also always be with Y) instead of a one-way condition. Another error is forgetting that X is prohibited whenever Y is present. Clearly separating the cases based on whether Y is in or out of the committee helps prevent such logical errors. Final Answer: The number of different committees that can be formed under the given conditions is 91.

Question 6

In how many different ways can the letters of the word HAPPYHOLI be arranged, taking all letters each time and considering that some letters are repeated?

Accepted Answer

Introduction / Context: This question involves counting permutations of letters in a word that contains repeated letters. It checks whether the student can correctly apply the permutation formula for multisets. Accurate counting of repeated letters is crucial to avoid overcounting arrangements that look the same because some letters are identical. Given Data / Assumptions: The word is HAPPYHOLI. The letters are H, A, P, P, Y, H, O, L, I. Total number of letters: 9. Repeated letters: H appears twice and P appears twice. We consider all arrangements of these 9 letters, counting only distinct permutations. Concept / Approach: If all letters were distinct, the total number of permutations would be 9!. However, because some letters are repeated, permuting identical letters among themselves does not create new distinct arrangements. To correct for this overcounting, we divide by the factorial of the number of times each letter is repeated. Therefore, the number of distinct permutations is 9! / (2! × 2!). Step-by-Step Solution: Step 1: Count total letters: there are 9 positions to fill, one for each letter of HAPPYHOLI.Step 2: Identify repetitions: the letter H appears 2 times, and the letter P appears 2 times. All other letters A, Y, O, L and I appear once.Step 3: Apply the permutation formula for repeated letters: total distinct arrangements = 9! / (2! × 2!).Step 4: Compute 9! = 362880.Step 5: Compute 2! × 2! = 2 × 2 = 4.Step 6: Divide 362880 by 4 to get 90720.Step 7: Therefore, the number of distinct arrangements of the letters of HAPPYHOLI is 90720. Verification / Alternative check: A quick reasonableness check is to compare with the case where all letters are distinct. That would give 362880 arrangements. Since two letters are repeated twice, we expect a reduction by a factor of 4 (2! for H and 2! for P), giving around 90000 arrangements. Our exact value 90720 fits this expectation perfectly, confirming the calculation. Why Other Options Are Wrong: Values like 72000 or 81000 correspond to different divisors or mistaken counts of repeated letters. 89772 or other similar numbers do not match 9! divided by any simple product of factorials of frequencies. Only 90720 equals 9! / (2! × 2!) and correctly reflects the presence of two repeated letters in the word. Common Pitfalls: Students sometimes miscount the number of times each letter appears, for example, noticing the two P's but missing that H also appears twice. Another common error is to use 9! directly without dividing by the factorials of repeated letters, thereby overcounting. Carefully writing out the letter frequencies before applying the formula is a good habit for such problems. Final Answer: The number of distinct arrangements of the letters of HAPPYHOLI is 90,720.

Question 7

The letters of the word PROMISE are to be arranged in all possible ways, but with the restriction that the three vowels in the word do not all come together as one block. How many different arrangements of the letters of PROMISE are possible under t…

Accepted Answer

Introduction / Context: This permutation problem involves a restriction on vowel placement. The word PROMISE contains three vowels, and we must count arrangements where these three vowels are not all together in one consecutive block. This type of question is classic in aptitude tests, where it is often easier to count the complement (arrangements with all vowels together) and subtract from the total number of unrestricted arrangements. Given Data / Assumptions: The word is PROMISE. Letters: P, R, O, M, I, S, E. Total letters: 7, all distinct. Vowels: O, I and E (three vowels). The condition is that the three vowels must not all come together as one consecutive block. Concept / Approach: Let us first count all possible arrangements of the 7 distinct letters without any restriction. Then we count the number of arrangements where the three vowels are together as a single block. Finally, we subtract this latter count from the total to obtain the number of arrangements where the three vowels are not all together. This complement approach simplifies the counting significantly. Step-by-Step Solution: Step 1: Count total arrangements without any restriction. Since there are 7 distinct letters, the total is 7! = 5040.Step 2: To count arrangements where all three vowels are together, treat the block of vowels (O, I, E) as one single unit.Step 3: The consonants are P, R, M and S, so along with the vowel block we have 5 units in total: [OIE], P, R, M and S.Step 4: The number of ways to arrange these 5 distinct units is 5!.Step 5: Compute 5! = 120.Step 6: Inside the vowel block, the three vowels O, I and E can be arranged among themselves in 3! ways.Step 7: Compute 3! = 6.Step 8: Total arrangements with all three vowels together equals 5! × 3! = 120 × 6 = 720.Step 9: Therefore, arrangements where the three vowels do not all come together as a single block equals total arrangements minus those with vowels together: 5040 - 720 = 4320. Verification / Alternative check: We can check reasonableness: arrangements with all vowels together are a subset of all permutations, so their count must be smaller than 5040. Having 720 such arrangements is plausible, leaving 4320 arrangements where the vowels are not all consecutively grouped. The complement method is standard and reliable for this type of constraint, reinforcing the correctness of the answer. Why Other Options Are Wrong: Numbers such as 4694, 4957 and 4871 do not arise from any simple factorial based decomposition of the problem. The total number of unrestricted arrangements is 5040, so no answer larger than this is possible. Among the options, only 4320 equals 5040 minus 720, where 720 is the correct count of arrangements with the vowels together as a block. Common Pitfalls: Some students misinterpret the condition and exclude arrangements where any two vowels are adjacent, which is not required. The question only forbids the three vowels from all being together; they can still be partly adjacent. Another common error is to forget to multiply 5! by 3! when counting the arrangements with the vowel block, leading to an incorrect count of 120 instead of 720 and hence an incorrect final answer. Final Answer: The number of arrangements of the letters of PROMISE in which the three vowels do not all come together is 4320.

Question 8

In how many distinct permutations can the letters of the word RAILINGS be arranged if the letters R and S are always together as a single adjacent block (in either order)?

Accepted Answer

Introduction / Context: This question tests permutations with a grouping restriction. We need to count how many different arrangements of the letters in the word RAILINGS are possible when the letters R and S must always appear next to each other, acting like a single block. This is a classic arrangement problem involving repeated letters and a special block condition. Given Data / Assumptions: The word is RAILINGS. Letters: R, A, I, L, I, N, G, S. The letter I appears twice, all other letters are distinct. R and S must always be adjacent, either as RS or SR. Concept / Approach: We treat the pair (R, S) as a single block for counting, and remember that this block itself can be internally arranged in 2 ways (RS or SR). Then we arrange this block along with the remaining letters, taking care of the repeated letter I. Step-by-Step Solution: Step 1: Write the multiset of letters: {R, A, I, I, L, N, G, S}. Step 2: Combine R and S into a single block X. Inside the block, the order can be RS or SR. Step 3: Now the objects to arrange are: X, A, I, I, L, N, G. That is 7 items with I repeated twice. Step 4: Number of arrangements of these 7 items = 7! / 2! because of the two identical I letters. Step 5: For each such arrangement, the internal order of R and S within block X can be RS or SR, so multiply by 2. Step 6: Total arrangements = (7! / 2!) * 2 = 7! = 5040. Verification / Alternative check: Another way is to directly note that 2 in the numerator cancels the 2 in the denominator, leaving 7!. Since 7! = 5040, the count is consistent and there is no double counting or omission. Why Other Options Are Wrong: 1260 and 2520 are smaller than 5040 and typically arise from dividing by extra factors that are not justified. 1080 is even smaller and does not match any meaningful factorial based count for this multiset and grouping condition. Common Pitfalls: Students often forget that R and S can be arranged in 2 ways inside their block or forget to divide by 2! for the repeated I. Some also mistakenly treat all letters as distinct, which gives 8! instead of the corrected grouped and repeated count. Final Answer: The number of permissible permutations of the letters of RAILINGS with R and S always together is 5040.

Question 9

A bag contains 8 red flowers, 7 blue flowers and 6 green flowers. If one flower is picked at random, what is the probability that the chosen flower is neither red nor green (that is, it is blue)?

Accepted Answer

Introduction / Context: This problem tests basic probability concepts in a simple selection setting. We have a mixture of differently coloured flowers in a bag and we pick one flower at random. The goal is to find the probability that the flower is neither red nor green, which effectively means we want the probability that the flower is blue. Given Data / Assumptions: Number of red flowers = 8. Number of blue flowers = 7. Number of green flowers = 6. One flower is selected at random. Each flower is equally likely to be selected. Concept / Approach: Probability of an event is defined as: favorable outcomes divided by total possible outcomes, provided all outcomes are equally likely. Here the favorable event is choosing a blue flower, because blue is the only colour that is neither red nor green. Step-by-Step Solution: Step 1: Compute the total number of flowers. Total flowers = 8 red + 7 blue + 6 green = 21 flowers. Step 2: Identify the favourable outcomes. The flower must be neither red nor green, so it must be blue, and there are 7 blue flowers. Step 3: Use the probability formula. Probability = favourable outcomes / total outcomes = 7 / 21. Step 4: Simplify the fraction 7 / 21 by dividing numerator and denominator by 7. Probability = 1 / 3. Verification / Alternative check: As a quick check, the probability of picking red is 8 / 21 and of picking green is 6 / 21. The probability of red or green is (8 + 6) / 21 = 14 / 21 = 2 / 3. Therefore, the probability of not red and not green is 1 minus 2 / 3 which is 1 / 3, confirming the result. Why Other Options Are Wrong: 7/20 and 3/7 use incorrect denominators or numerators that do not match the total counts. 2/3 corresponds to picking red or green, which is the complement event rather than the desired one. Common Pitfalls: A common mistake is to forget to include all flower types when computing the total or to misinterpret the phrase neither red nor green. Another error is to subtract probabilities incorrectly instead of using the direct count of blue flowers. Final Answer: The probability that the randomly selected flower is neither red nor green is 1/3.

Question 10

From a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can this selection be made such that at least one boy is included in the group of four?

Accepted Answer

Introduction / Context: This question checks understanding of combinations and the use of complementary counting. We must choose a group of four children from a mixed group of boys and girls, under the condition that there is at least one boy in the selected group. Given Data / Assumptions: Total boys = 6. Total girls = 4. We select 4 children in total. The selection should contain at least one boy. Concept / Approach: The simplest way is to first count all possible groups of four children without any restriction, and then subtract the number of groups that violate the condition. The only disallowed case is a group with no boys at all, that is, a group containing only girls. Step-by-Step Solution: Step 1: Total number of children = 6 + 4 = 10. Step 2: Total ways to choose any 4 children from 10 = C(10, 4). Step 3: Evaluate C(10, 4) = 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1) = 210. Step 4: Now count the groups with no boys (only girls). We have 4 girls and need to select 4 of them. Step 5: Number of all-girl groups = C(4, 4) = 1. Step 6: Subtract these invalid groups from the total. Valid groups with at least one boy = 210 - 1 = 209. Verification / Alternative check: You could also count directly by cases (for example, 1 boy and 3 girls, 2 boys and 2 girls, 3 boys and 1 girl, and 4 boys and 0 girls) and add the results. Doing this carefully will also yield 209, which confirms the complementary counting method. Why Other Options Are Wrong: 205, 194 and 159 arise if some valid distributions are omitted or if an incorrect total number of combinations is used. They do not match the value obtained either by complementary counting or by the casewise approach. Common Pitfalls: Many learners forget to subtract the all-girl case or accidentally subtract more cases than necessary. Others confuse combinations with permutations and incorrectly multiply by extra factors based on ordering, which is not relevant here since the order of children in the group does not matter. Final Answer: The number of ways to select four children with at least one boy present is 209.

Question 11

In how many distinct permutations can the letters of the word TRANSFORMER be arranged if the letters N and S always appear together as adjacent letters (in either order)?

Accepted Answer

Introduction / Context: This question involves permutations of a word that has repeated letters and an additional restriction that two specified letters must always appear together. The word is TRANSFORMER and we are asked to count arrangements where N and S are adjacent in the sequence. Given Data / Assumptions: Word: TRANSFORMER. Letters: T, R, A, N, S, F, O, R, M, E, R. The letter R appears three times; all other letters appear once. The letters N and S must always be together as NS or SN. Concept / Approach: We treat the pair N and S as a single block that can internally be arranged in 2 ways (NS or SN). Then we arrange this block together with the remaining letters, taking into account that R is repeated three times. This is a standard block and repetition problem in permutations. Step-by-Step Solution: Step 1: Remove N and S from the list and treat them as a block X. Step 2: The remaining letters are T, R, A, F, O, R, M, E, R and the block X. That gives a total of 10 objects. Step 3: Among these 10 objects, the letter R occurs three times, while all other objects are distinct. Step 4: Number of permutations of 10 objects with 3 identical R letters is 10! / 3!. Step 5: Inside the block X, the pair can be arranged as NS or SN, giving 2 internal arrangements. Step 6: Total arrangements = (10! / 3!) * 2. Step 7: Compute the value: 10! = 3628800 and 3! = 6, so 10! / 3! = 604800. Multiply by 2 to get 1209600. Verification / Alternative check: You can double check the count by confirming the multiset structure and ensuring that only the three R letters are considered identical. No further division is needed because all other letters and the NS block are distinct. The final multiplication by 2 accounts correctly for both NS and SN orderings. Why Other Options Are Wrong: 112420, 85120 and 40320 are all significantly smaller than the correct figure and typically come from ignoring either the repeated R letters or the block constraint, or from miscounting the total number of objects to arrange. Common Pitfalls: Students may forget to divide by 3! for the three R letters or fail to multiply by 2 for the internal arrangement of N and S. Others may mistakenly treat N and S as always in the same internal order, which halves the correct count. Final Answer: The required number of permutations of TRANSFORMER with N and S always together is 1209600.

Question 12

Evaluate the permutation 100P2, that is, find the numerical value of the number of permutations of 2 objects chosen from 100 distinct objects.

Accepted Answer

Introduction / Context: The question asks for the value of a permutation expression, 100P2. This notation represents the number of ways to arrange 2 distinct items chosen from a set of 100 distinct items, where the order of the chosen items matters. Given Data / Assumptions: The total number of distinct objects, n, is 100. The number of positions to be filled, r, is 2. We use the permutation formula nPr for arrangements where order is important. Concept / Approach: The formula for permutations is nPr = n * (n - 1) * (n - 2) ... up to r factors. For small r, this simplifies to a product of r consecutive integers starting from n and decreasing by 1 each time. For 100P2, we only need two factors: 100 and 99. Step-by-Step Solution: Step 1: Recall the formula for permutations. nPr = n! / (n - r)!. Step 2: Substitute n = 100 and r = 2. 100P2 = 100! / 98!. Step 3: Cancel common terms in the factorials. 100! / 98! = 100 * 99 because all the factors from 98 downwards cancel out. Step 4: Compute the product 100 * 99. 100 * 99 = 9900. Verification / Alternative check: You can also think of this as choosing the first position in 100 ways and the second position in 99 ways. Since each ordered pair is distinct, the total number of ordered pairs is 100 * 99, which again gives 9900. Why Other Options Are Wrong: 10000 would correspond to 100 * 100, which ignores the fact that once one element is chosen, it cannot be chosen again. 8900 and 7900 are arbitrary numbers that do not arise from the correct permutation formula or any logical counting method here. Common Pitfalls: A frequent mistake is to mix up permutations and combinations and try to use nCr instead. Another error is to allow repetition, which would lead to 100^2, not the required product 100 * 99. Final Answer: The value of the permutation 100P2 is 9900.

Question 13

In the word TROUBLED, how many pairs of letters are there such that the number of letters between them in the word is the same as the number of letters between them in the English alphabet?

Accepted Answer

Introduction / Context: This question tests careful counting and understanding of alphabetical positions. We are asked to find pairs of letters in the word TROUBLED where the gap between the letters in the word matches the gap between the same letters in the English alphabet, measured in terms of letters in between. Given Data / Assumptions: Word: TROUBLED. Letters in order: T, R, O, U, B, L, E, D. Alphabetical positions are based on A = 1, B = 2, and so on up to Z = 26. For a pair of letters, we compare the number of letters between them in the word and in the alphabet. Concept / Approach: For any pair of letters at positions i and j in the word (i less than j), the number of letters between them in the word is (j minus i minus 1). In the alphabet, if their positions are p1 and p2, the number of letters between them is the absolute value of (p1 minus p2) minus 1. We need to find pairs where these two counts are equal. Step-by-Step Solution: Step 1: Write the word with indices: T(1), R(2), O(3), U(4), B(5), L(6), E(7), D(8). Step 2: For each pair of positions, compute the word gap j - i - 1. Step 3: Map each letter to its alphabet position (for example, A = 1, B = 2, ..., Z = 26). For O and L, O is 15, L is 12 and so on. Step 4: For each pair, compute the alphabet gap abs(p1 - p2) - 1. Step 5: Check equality of the word gap and alphabet gap. Step 6: The pairs that satisfy this condition are O and L, and E and D. Verification / Alternative check: You can verify individual pairs explicitly. For example, O is at position 3 in the word and L is at position 6, so there are 6 - 3 - 1 = 2 letters between them in TROUBLED. In the alphabet, O is letter 15 and L is letter 12, so there are abs(15 - 12) - 1 = 2 letters between them as well. A similar verification works for E and D. Why Other Options Are Wrong: Options 3, 4 and 5 assume more matching pairs than exist. A detailed check shows only two valid pairs satisfy the equal gap condition, so those larger counts are overestimates. Common Pitfalls: Many students miscount letters between positions or forget to subtract one when converting positions into counts of letters in between. Another common issue is to miscalculate alphabetical positions for letters, especially later in the alphabet. Final Answer: The number of such pairs of letters in TROUBLED is 2.

Question 14

Eight people A, B, C, D, E, F, G and H sit around a circular table facing the centre (not necessarily in the same order). G sits third to the right of C. E is second to the right of G and fourth to the right of H. B is fourth to the right of C. D is…

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Introduction / Context: This is a circular seating arrangement problem that tests logical reasoning, spatial arrangement skills and the consistent application of multiple conditions. Eight people, all facing the centre, must be seated in such a way that all given constraints on relative positions are satisfied. Then we must identify which statement about F and B and about B and D is correct. Given Data / Assumptions: People: A, B, C, D, E, F, G, H. All sit around a circular table facing the centre. G is third to the right of C. E is second to the right of G and fourth to the right of H. B is fourth to the right of C. D is not an immediate neighbour of E. A and C are immediate neighbours. Concept / Approach: We place one person at a reference position (for example, C) and then place others step by step using the right direction consistently around the circle. Because everyone faces the centre, moving to the right corresponds to moving clockwise. Once a valid arrangement is built, we can read off the relative positions needed to evaluate the answer choices. Step-by-Step Solution: Step 1: Fix C at a reference seat. Place G third to the right of C. Step 2: Place B fourth to the right of C. Step 3: Place E second to the right of G and then place H so that E is fourth to the right of H. Step 4: Ensure that A sits adjacent to C, as an immediate neighbour. Step 5: Place D and F in the remaining seats while respecting the condition that D is not a neighbour of E. Step 6: One valid final circular ordering (clockwise) is A, C, H, D, G, B, E, F. Verification / Alternative check: From this arrangement, starting from B and moving clockwise, the first seat to the right is E and the second seat to the right is F. Therefore, F is second to the right of B, which makes option stating this fact correct. F is not third to the left of B and B is not next to D, so those statements are false. Why Other Options Are Wrong: The statement that F is third to the left of B does not match the derived seating order. The statement that B is an immediate neighbour of D is also not supported by any consistent arrangement satisfying all conditions. Therefore, the option claiming all of the statements are true is incorrect as well. Common Pitfalls: Learners often mix up left and right in circular arrangements or change direction midway. Another issue is not maintaining a fixed orientation when rotating the circle, which can lead to inconsistent conclusions about relative positions. Final Answer: The only correct statement is that F is second to the right of B.

Question 15

You have 6 different subject books: English, Hindi, Mathematics, History, Geography and Science. In how many distinct ways can these 6 books be arranged on a single shelf?

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Introduction / Context: This is a basic permutation question involving the arrangement of distinct objects in a row. Each subject book is considered different, and we are asked to find how many different linear arrangements are possible on a single shelf. Given Data / Assumptions: Books: English, Hindi, Mathematics, History, Geography, Science. Total number of books = 6. All books are distinct. Order of books on the shelf matters. Concept / Approach: When arranging n distinct items in a row, the number of possible permutations is n factorial, written as n!. This is because we have n choices for the first position, n - 1 choices for the second position, and so on until the last position. Step-by-Step Solution: Step 1: Identify that there are 6 distinct books. Step 2: The number of ways to arrange 6 distinct items in a row is 6!. Step 3: Evaluate 6! = 6 * 5 * 4 * 3 * 2 * 1. Step 4: Compute the product: 6 * 5 = 30, 30 * 4 = 120, 120 * 3 = 360, 360 * 2 = 720, and 720 * 1 = 720. Step 5: Therefore, there are 720 distinct permutations. Verification / Alternative check: You can think sequentially: there are 6 choices for the first place, 5 remaining for the second place, 4 for the third, 3 for the fourth, 2 for the fifth and 1 for the last. Multiplying these together gives 6 * 5 * 4 * 3 * 2 * 1, which again is 720, confirming the factorial result. Why Other Options Are Wrong: 360 equals 6! divided by 2 and would arise if some incorrect restriction or symmetry were applied. 240 and 780 do not correspond to any standard permutation count of 6 distinct items and result from miscalculations or wrong formula usage. Common Pitfalls: Some learners confuse permutations with combinations and try to use nCr instead of n!. Others may undercount by forgetting to include all positions in the factorial or overcomplicate a straightforward arrangement problem. Final Answer: The six different subject books can be arranged on the shelf in 720 distinct ways.

Question 16

Using each of the digits 1, 3, 5 and 7 exactly once in each number, how many 4 digit numbers can be formed in total, and what is the sum of all such 4 digit numbers?

Accepted Answer

Introduction / Context: This problem combines permutations with place value concepts. We are asked to form all possible 4 digit numbers using the digits 1, 3, 5 and 7, with each digit used exactly once in each number, and then to compute the sum of all such numbers. Given Data / Assumptions: Digits available: 1, 3, 5, 7. Each 4 digit number must use each digit exactly once. No repetition of digits within a single number. All permutations of these four digits are valid 4 digit numbers. Concept / Approach: The number of distinct 4 digit numbers formed from these four digits is 4!, since all digits are distinct. Each digit appears an equal number of times in each place value (units, tens, hundreds, thousands). We can exploit this symmetry to compute the total sum without listing every number. Step-by-Step Solution: Step 1: Count the total number of permutations: 4! = 4 * 3 * 2 * 1 = 24. Step 2: Each of the four digits appears equally often in each place. In 24 permutations, each digit appears 24 / 4 = 6 times in any given position. Step 3: Sum of the digits = 1 + 3 + 5 + 7 = 16. Step 4: Contribution of the thousands place = 6 * 16 * 1000. Step 5: Contribution of the hundreds place = 6 * 16 * 100. Step 6: Contribution of the tens place = 6 * 16 * 10. Step 7: Contribution of the units place = 6 * 16 * 1. Step 8: Factor out 6 * 16 = 96. Total sum = 96 * (1000 + 100 + 10 + 1) = 96 * 1111. Step 9: Compute 96 * 1111 = 106656. Verification / Alternative check: As a check, you could compute the sum for a smaller example with two or three digits and verify the pattern. The symmetry argument that each digit appears the same number of times in each position remains valid and scales correctly to this case as well. Why Other Options Are Wrong: 105555, 106665 and 108333 are near but not equal to the computed total. They reflect small arithmetic mistakes or incorrectly grouped contributions from place values. Common Pitfalls: Mistakes often occur in counting how many times each digit appears in each place or in adding up place value contributions. Some students attempt to list all 24 numbers manually, which is time consuming and prone to error. Final Answer: The sum of all 4 digit numbers formed using 1, 3, 5 and 7 exactly once in each number is 106656.

Question 17

If repetition of digits is allowed, how many three digit even natural numbers can be formed in total?

Accepted Answer

Introduction / Context: This question tests understanding of counting principles for number formation. We want to count the number of three digit even natural numbers when digits can be repeated. The even condition restricts the last digit, while the three digit condition restricts the first digit. Given Data / Assumptions: Digits available: 0 to 9. We are forming three digit natural numbers. The numbers must be even. Repetition of digits is allowed. Concept / Approach: A three digit number has a hundreds, tens and units place. For a valid three digit number, the hundreds digit cannot be 0. For an even number, the units digit must be one of the even digits 0, 2, 4, 6 or 8. The tens digit can be any digit from 0 to 9 because repetition is allowed. Step-by-Step Solution: Step 1: Determine the choices for the units digit (last digit). To be even, it must be one of 0, 2, 4, 6 or 8. Step 2: The number of choices for the units digit is 5. Step 3: The hundreds digit (first digit) cannot be 0, so it can be any of 1, 2, 3, 4, 5, 6, 7, 8 or 9. Step 4: The number of choices for the hundreds digit is 9. Step 5: The tens digit has no restriction other than being a digit from 0 to 9, and repetition of digits is allowed. Step 6: The number of choices for the tens digit is 10. Step 7: Multiply the independent choices: total even three digit numbers = 9 * 10 * 5 = 450. Verification / Alternative check: You can reason by cases based on the last digit. For each fixed even last digit, the hundreds digit has 9 choices and the tens digit has 10 choices, giving 9 * 10 = 90 numbers per last digit. With 5 possible last digits, the total is 5 * 90 = 450, which matches the earlier product. Why Other Options Are Wrong: 500, 540 and 550 correspond to miscounts, often from allowing 0 as a hundreds digit, misunderstanding the even condition, or miscounting the number of even digits available for the last position. Common Pitfalls: A major mistake is allowing the first digit to be 0, which leads to two digit or smaller numbers. Another mistake is to think that there are only four even digits by incorrectly excluding 0 from the units place, even though it is allowed for the last position. Final Answer: The total number of three digit even natural numbers with repetition allowed is 450.

Question 18

In how many different permutations can the letters of the word THERAPY be arranged such that the two vowels E and A never appear together as adjacent letters?

Accepted Answer

Introduction / Context: This problem is about permutations with a restriction that certain letters must not be adjacent. The word THERAPY has two vowels, and we want to count the number of arrangements in which these vowels do not come together as a single block. Given Data / Assumptions: Word: THERAPY. Letters: T, H, E, R, A, P, Y. Total letters = 7 and all are distinct. Vowels: E and A; consonants: T, H, R, P, Y. Vowels must not be adjacent in the arrangement. Concept / Approach: A standard strategy is to first count all possible permutations without any restriction, and then subtract the number of permutations where the vowels are together as a single block. The result gives the arrangements where the vowels never come together. Step-by-Step Solution: Step 1: Total permutations of 7 distinct letters without restriction = 7!. Step 2: Compute 7! = 5040. Step 3: Now count permutations where E and A are together. Treat the pair (E, A) or (A, E) as a single block X. Step 4: Then the objects to arrange are X and the 5 consonants T, H, R, P, Y. So we have 6 distinct objects. Step 5: Arrangements of these 6 distinct objects = 6! = 720. Step 6: Inside the block X, E and A can be arranged as EA or AE, giving 2 internal arrangements. Step 7: Total arrangements with vowels together = 6! * 2 = 720 * 2 = 1440. Step 8: Arrangements where vowels never come together = total permutations minus permutations with vowels together = 5040 - 1440 = 3600. Verification / Alternative check: You can also think of placing the consonants first and then inserting vowels in the gaps, but the subtraction method is straightforward and reliable for this case. The calculation 5040 - 1440 clearly leads to 3600 as the count of valid arrangements. Why Other Options Are Wrong: 1440 is the count of arrangements where the vowels are together, not where they are separated. 720 is 6!, which is insufficient here. 2250 is not linked to any standard counting step in this approach. Common Pitfalls: The main error is to count only the cases where vowels are together and forget to subtract from the total. Some also forget the factor of 2 for the internal arrangements of the vowels within the block. Final Answer: The number of permutations of THERAPY in which the vowels never appear together is 3600.

Question 19

In how many permutations can the letters of the word NUMERICAL be arranged if all consonants are required to occupy only the even positions? If such an arrangement is not possible, what should the correct count be?

Accepted Answer

Introduction / Context: This question appears to ask for the number of permutations of the letters of the word NUMERICAL under the restriction that all consonants must occupy even positions in the arrangement. Analysing whether this is even feasible is a key part of solving the problem correctly. Given Data / Assumptions: Word: NUMERICAL. Letters: N, U, M, E, R, I, C, A, L. Total letters = 9, so positions are 1 to 9. Vowels: U, E, I, A (4 in total). Consonants: N, M, R, C, L (5 in total). Consonants are required to occupy even positions only, namely positions 2, 4, 6 and 8. Concept / Approach: Even positions available are 2, 4, 6 and 8, which provide only 4 slots. However, there are 5 consonants. If every consonant must be in an even position, we would need at least 5 even slots, which is not the case for a 9 letter word. Therefore, the arrangement is impossible and the correct count of valid permutations is zero. Step-by-Step Solution: Step 1: List the consonants: N, M, R, C, L, so we have 5 consonants. Step 2: List the available even positions: 2, 4, 6, 8. This provides only 4 positions. Step 3: Since each consonant must occupy an even position, but there are more consonants than even positions, it is impossible to place all 5 consonants without violating the restriction. Step 4: Because the fundamental requirement cannot be met, there is no valid arrangement of the letters. Step 5: Therefore, the total number of valid permutations is 0. Verification / Alternative check: Even if you attempt to fill the even positions first, you will soon see that one consonant always remains without an available even slot. There is no way to move a consonant to an odd position because that would break the restriction, so the impossibility is clear. Why Other Options Are Wrong: Values such as 120, 720 or 24 come from permutations of smaller subsets or from ignoring the mismatch between the number of consonants and the number of even positions. They do not reflect the feasibility constraint. Common Pitfalls: A typical error is to rush into factorial calculations without first checking whether the arrangement requirement is even possible. Counting formulas must always respect feasibility conditions like the number of available slots. Final Answer: Since there are more consonants than available even positions, no valid arrangement exists and the correct count of such permutations is 0.

Question 20

Five men and four women are to be seated in a row such that all the four women occupy only the even numbered positions in the row. How many such linear arrangements are possible?

Accepted Answer

Introduction / Context: This arrangement question combines permutations with a positional restriction. We have five men and four women to be seated in a row, with the condition that every woman must occupy an even position in the row. Given Data / Assumptions: Total people = 5 men + 4 women = 9. Positions in the row are numbered 1 to 9 from left to right. Women must sit only in even positions (2, 4, 6 and 8). All men and women are distinct individuals. Concept / Approach: We treat seating as a two stage process. First, we seat the four women in the four available even positions. Then we seat the five men in the remaining five positions (all of which are odd positions). Since all individuals are distinct, we use factorial counts for each group and multiply the results. Step-by-Step Solution: Step 1: Identify the even positions in the row: 2, 4, 6 and 8. There are 4 such positions. Step 2: The four women must occupy these four even positions. Step 3: Number of ways to arrange 4 distinct women in 4 distinct positions = 4! = 24. Step 4: The remaining positions are 1, 3, 5, 7 and 9, which are 5 positions to be filled by the 5 men. Step 5: Number of ways to arrange 5 distinct men in 5 positions = 5! = 120. Step 6: Since choices are independent, total arrangements = 4! * 5! = 24 * 120 = 2880. Verification / Alternative check: You could reverse the order by first placing men in the 5 odd positions and then women in the even positions, but the final multiplication remains 5! * 4! which is 2880. This confirms the earlier calculation and shows that the order of the two stages does not affect the final count. Why Other Options Are Wrong: 1440 and 720 correspond to using either 4! alone or 5! alone or dividing incorrectly. 2020 does not correspond to any natural factorial combination here and reflects a miscalculation. Common Pitfalls: Errors often arise from treating the problem as if there were fewer restrictions or from assuming that some positions are indistinguishable. Another common issue is accidentally swapping counts for men and women or missing one group in the multiplication. Final Answer: The number of valid linear arrangements is 2880.

Permutation and Combination Questions