Question 1

From a bunch of flowers containing 16 red roses and 14 white roses (30 roses in total), four flowers have to be selected. In how many different ways can this selection be made if at least one of the selected flowers is a red rose?

Accepted Answer

Introduction / Context: This problem asks you to choose 4 flowers from a mixture of red and white roses, under the condition that at least one of the chosen flowers must be red. It is a classic example of using combinations with a simple condition and is best approached using the complement method: count all possible selections and then subtract the selections that violate the condition (that is, selections with no red roses at all). This method simplifies the counting compared to enumerating many separate subcases. Given Data / Assumptions: - Number of red roses: 16. - Number of white roses: 14. - Total roses: 16 + 14 = 30. - We must select exactly 4 flowers. - Condition: at least one selected flower must be a red rose. - Flowers of the same colour are distinct in counting, and order of selection does not matter; only the chosen subset counts. Concept / Approach: Let us first count the total number of ways to select any 4 flowers from the 30 available roses, without any colour restriction. Then we count how many of those selections contain no red rose (that is, all 4 flowers are white). Selections that meet the "at least one red" condition are all selections minus those with zero red roses. This is the complement approach and is often much easier than splitting into multiple red-white combinations for 1 red, 2 reds, 3 reds and 4 reds. Step-by-Step Solution: Step 1: Compute the total number of ways to choose 4 flowers from 30 roses.Step 2: Total selections (no restriction) = 30C4.Step 3: Compute 30C4 = (30 * 29 * 28 * 27) / (4 * 3 * 2 * 1) = 27405.Step 4: Next, count the number of selections with no red roses, that is, all 4 flowers are white.Step 5: There are 14 white roses, so such selections are counted by 14C4.Step 6: Compute 14C4 = (14 * 13 * 12 * 11) / (4 * 3 * 2 * 1) = 1001.Step 7: Selections with at least one red rose = total selections - selections with zero red roses.Step 8: Therefore, required count = 30C4 - 14C4 = 27405 - 1001 = 26404. Verification / Alternative check: As a check, we can add up the cases by number of red roses directly. Let r be the number of reds selected (r ranges from 1 to 4). The count becomes sum over r of 16Cr * 14C(4-r). Computing explicitly: for r = 1, we have 16C1 * 14C3 = 16 * 364 = 5824; for r = 2, 16C2 * 14C2 = 120 * 91 = 10920; for r = 3, 16C3 * 14C1 = 560 * 14 = 7840; for r = 4, 16C4 * 14C0 = 1820 * 1 = 1820. Adding these: 5824 + 10920 + 7840 + 1820 = 26404, which matches the complement method result. Why Other Options Are Wrong: - 27405 counts all selections with no colour restriction, including those with zero red roses, so it does not enforce the "at least one red" condition. - 26584 and 26585 are near the correct value but arise from miscalculating one of the combinations or making an arithmetic slip in subtraction. Common Pitfalls: Many learners attempt to split the problem into many cases (exactly 1 red, 2 reds, 3 reds, 4 reds) and forget one of the cases or miscalculate a term. Others misinterpret "at least one red" as "exactly one red". The complement method is safer and usually quicker: total minus "no red". Additionally, errors often occur in computing large combinations such as 30C4; doing the multiplication and division carefully or cancelling common factors early reduces mistakes. Final Answer: The number of ways to select 4 flowers so that at least one is a red rose is 26404.

Question 2

A single card is drawn at random from a standard pack of 52 playing cards. What is the probability that the card drawn is either black or a king (or both)?

Accepted Answer

Introduction / Context: This is a probability question involving a standard deck of playing cards and a union of two events: drawing a black card or drawing a king. Such problems are a good test of understanding of the addition rule for probabilities, especially the need to subtract the overlap when events are not mutually exclusive. The classical deck has 52 cards divided evenly by colour and suit, and kings are a well-known subset in each suit. Given Data / Assumptions: - Standard deck: 52 cards in total. - There are 26 black cards (13 spades and 13 clubs). - There are 4 kings in total (one in each suit: hearts, diamonds, clubs, spades). - Among these 4 kings, 2 are black (king of spades and king of clubs) and 2 are red (king of hearts and king of diamonds). - We want P(black or king), where "or" is inclusive: black, king, or both. Concept / Approach: We use the addition rule for probabilities of two events A and B: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Here, let A be the event "card is black", and B be the event "card is a king". We first count how many cards satisfy each individual event, then count how many satisfy both (black kings), and finally substitute into the formula to avoid double-counting the intersection. Step-by-Step Solution: Step 1: Count black cards: there are 26 black cards out of 52.Step 2: So P(A) = 26/52.Step 3: Count kings: there are 4 kings out of 52 cards.Step 4: So P(B) = 4/52.Step 5: Count black kings: there are 2 black kings (clubs and spades).Step 6: So P(A ∩ B) = 2/52.Step 7: Apply the addition rule: P(A ∪ B) = 26/52 + 4/52 - 2/52.Step 8: Combine the fractions: (26 + 4 - 2) / 52 = 28/52.Step 9: Simplify 28/52 by dividing numerator and denominator by 4: 28/52 = 7/13. Verification / Alternative check: We can also count favourable cards directly. A card is favourable if it is black (26 cards) and not a king (26 - 2 = 24 cards), or is a king of any colour (4 cards). So total favourable = 24 non-king black cards + 4 kings = 28 cards. Dividing by 52 gives 28/52 = 7/13, which matches our earlier calculation using the addition rule. This double check confirms both counting methods agree. Why Other Options Are Wrong: - 15/26 corresponds to (26 + 4) / 52 without subtracting the overlap of 2 black kings, so it double-counts those cards. - 15/52 and 17/26 do not arise from any correct application of the counting or probability formulas for this problem; they reflect miscounting or mis-simplified fractions. Common Pitfalls: A common mistake is forgetting to subtract P(A ∩ B) when events overlap, leading to double-counting the black kings. Another error is to misinterpret "either black or a king" as exclusive or, which would require subtracting the intersection instead of adding and then subtracting. Here, standard probability language uses inclusive or, so we must include black kings exactly once in the count. Final Answer: The probability that the card drawn is either black or a king is 7/13.

Question 3

In how many different ways can the letters of the word 'RITUAL' be arranged, assuming that all letters are used each time?

Accepted Answer

Introduction / Context: This is a basic permutation problem involving all the letters of a word, with no additional restrictions. The word RITUAL consists of six distinct letters, and we want to count all possible different arrangements (permutations) of these letters. This type of question reinforces the simple factorial-based formula for permutations of distinct objects and forms the foundation for more complex arrangement problems. Given Data / Assumptions: - Word: RITUAL. - Total letters: 6 (R, I, T, U, A, L), all distinct. - Each arrangement (word) uses all 6 letters exactly once. - Order of letters matters: different orders represent different arrangements. - There are no repeated letters and no positional restrictions. Concept / Approach: When arranging n distinct items in a sequence, the number of different permutations is n! (n factorial). This comes from the fact that there are n choices for the first position, (n - 1) choices for the second, and so on, down to 1 choice for the last position. Multiplying these choices gives n! possible arrangements. Here, n = 6, so we simply compute 6!. Step-by-Step Solution: Step 1: Count the total number of letters in the word RITUAL.Step 2: There are 6 distinct letters.Step 3: The number of permutations of 6 distinct letters is 6!.Step 4: Compute 6! = 6 * 5 * 4 * 3 * 2 * 1.Step 5: Multiply stepwise: 6 * 5 = 30; 30 * 4 = 120; 120 * 3 = 360; 360 * 2 = 720; 720 * 1 = 720.Step 6: Therefore, there are 720 different arrangements. Verification / Alternative check: You can think of it position-wise: for the first position, any of the 6 letters can be used. After choosing the first letter, 5 letters remain for the second position, 4 for the third, and so on, down to 1 for the last position. Thus, the total number of sequences is 6 * 5 * 4 * 3 * 2 * 1, which again equals 720. This confirms that the factorial formula matches the intuitive counting process. Why Other Options Are Wrong: - 5040 equals 7!, which would be the number of permutations of 7 distinct items; this is not applicable here because RITUAL has only 6 letters. - 360 is 6! / 2 and might arise if someone mistakenly divides by 2, perhaps thinking of symmetry or repetition where none exists. - 180 is even smaller and could result from prematurely stopping the multiplication or accidentally dividing by additional factors. Common Pitfalls: Some learners miscount the letters or incorrectly assume there is a repeated letter in the word, which would require dividing by factorials of repeat counts. Others simply confuse 5! and 6!, underestimating the total. Being careful to identify the exact number of distinct symbols and using the correct factorial is crucial. Final Answer: The letters of the word RITUAL can be arranged in 720 different ways.

Question 4

A committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee must include at least one lady?

Accepted Answer

Introduction / Context: This question deals with forming a committee from a mixed group of gentlemen and ladies with a minimum representation constraint. We have to form a committee of 5 people from a group of 10 (6 gentlemen and 4 ladies), ensuring that at least one lady is included. This is a good example of applying combinations and using the complement method to handle an "at least one" condition efficiently. Given Data / Assumptions: - Gentlemen available: 6. - Ladies available: 4. - Total people: 6 + 4 = 10. - Committee size: 5 members. - Condition: the committee must contain at least one lady (1 or more). - Order of members in the committee does not matter; we only care about the chosen set. Concept / Approach: We can count the number of valid committees by subtracting from the total number of possible committees (with no restrictions) the number of committees that violate the condition (that is, committees with zero ladies). The complement method is usually simpler than adding up cases for exactly 1 lady, exactly 2 ladies, and so on, which would require multiple combination terms. Step-by-Step Solution: Step 1: Compute the total number of ways to choose any 5 persons from the 10 available people, without any condition.Step 2: Total committees without restriction = 10C5.Step 3: Compute 10C5 = 252 (a standard value or via 10! / (5! * 5!) ).Step 4: Now count the number of committees with no ladies at all (only gentlemen).Step 5: Such committees must be formed entirely from the 6 gentlemen.Step 6: Number of all-gentlemen committees = 6C5 = 6.Step 7: Committees that contain at least one lady = total committees - all-gentlemen committees.Step 8: Required count = 10C5 - 6C5 = 252 - 6 = 246. Verification / Alternative check: As an alternative, we can sum over cases where the committee has exactly k ladies for k = 1, 2, 3, 4. For example, for exactly 1 lady, we choose 1 from 4 and 4 from 6 gentlemen; for exactly 2 ladies, choose 2 from 4 and 3 from 6, and so on. Adding all these cases yields the same result of 246, but it requires more arithmetic. The complement method is cleaner and less error-prone, and the equality of the two results verifies correctness. Why Other Options Are Wrong: - 123 and 113 are far too small and come from miscomputing either 10C5 or 6C5, or from trying to sum only some of the lady-count cases. - 945 is much larger than 252 and cannot be correct for choosing 5 people out of 10; it reflects a serious misunderstanding of the combination formula. Common Pitfalls: A frequent error is to forget that committees with no ladies must be excluded and to simply compute 10C5. Another pitfall is incorrectly summing several overlapping cases for different numbers of ladies, leading to double-counting or missing some configurations. Carefully applying the complement method—total minus invalid—is the simplest strategy in most “at least one” problems. Final Answer: The committee can be formed in 246 different ways if it must include at least one lady.

Question 5

In how many different ways can the letters of the word 'DESIGN' be arranged so that no consonant appears in either the first or the last position (that is, the two end positions must both be vowels)?

Accepted Answer

Introduction / Context: This problem involves arranging letters of a word with restrictions on the end positions. The word DESIGN consists of six distinct letters and a mix of vowels and consonants. The condition is that no consonant may appear at either end of the word. Therefore, both the first and last positions must be occupied by vowels. The remaining positions can be filled by consonants without any extra restrictions. This tests your ability to separate positional constraints from the general permutation. Given Data / Assumptions: - Word: DESIGN. - Letters: D, E, S, I, G, N (6 distinct letters). - Vowels: E, I (2 vowels). - Consonants: D, S, G, N (4 consonants). - Positions: 6, labelled from 1 to 6. - Condition: positions 1 and 6 (the two ends) must not be consonants, so they must be vowels. Concept / Approach: Because there are only two vowels and exactly two end positions, both vowels must occupy the first and last positions in some order. Once we place E and I in positions 1 and 6, the remaining 4 positions in the middle can be filled freely by the 4 consonants D, S, G, N. Since all letters are distinct, this becomes a straightforward multiplication of the permutations for vowels at the ends and permutations for consonants in the middle positions. Step-by-Step Solution: Step 1: Identify which letters are vowels and which are consonants.Step 2: Vowels: E and I; consonants: D, S, G, N.Step 3: The two end positions (first and last) must be occupied by vowels.Step 4: There are 2 vowels and 2 end positions, so the vowels can be arranged in these positions in 2! ways (E at first and I at last, or I at first and E at last).Step 5: The remaining 4 positions in the middle must be filled with the 4 consonants D, S, G, N.Step 6: All 4 consonants are distinct, so they can be arranged in the 4 middle positions in 4! ways.Step 7: Compute 4! = 4 * 3 * 2 * 1 = 24.Step 8: Total valid arrangements = (ways to arrange vowels at the ends) * (ways to arrange consonants in the middle) = 2! * 4! = 2 * 24 = 48. Verification / Alternative check: We can confirm that no other pattern is possible because there are exactly as many vowels as there are end slots. If one of the end positions were filled by a consonant, we would have to place at least one vowel in the middle, violating the condition that no consonant appears at either end. Therefore, the only valid structure is vowel–middle consonants–vowel, which our counting fully captures. Recomputing 2 * 24 also confirms the arithmetic as 48. Why Other Options Are Wrong: - 240 and 72 are larger than 48 and likely come from ignoring one of the constraints, such as treating the end positions as free or not separating vowel and consonant roles correctly. - 36 is smaller than 48 and might result from mistakenly dividing by 2 or 4, perhaps due to confusion over repeated letters (even though there are none here). Common Pitfalls: A common mistake is misidentifying vowels and consonants or assuming that vowels can also appear in the middle with some unused at the ends. But since only vowels can occupy the ends and there are precisely two vowels, all vowel positions are forced. Another error is forgetting that the order of consonants in the middle matters, leading to an undercount by using combinations instead of permutations. Final Answer: The letters of the word DESIGN can be arranged in 48 ways so that no consonant appears at either end.

Question 6

In how many different ways can the letters of the word 'DETAIL' be arranged such that all the vowels occupy only the odd positions (1st, 3rd and 5th positions)?

Accepted Answer

Introduction / Context: This problem is another arrangement question with positional restrictions, similar to other vowel-consonant placement problems. The word DETAIL has six letters, including three vowels and three consonants. The condition is that the vowels must occupy only the odd positions in the arrangement. Since there are exactly three odd positions in a six-letter word and exactly three vowels, each odd position must be occupied by a vowel, and each even position by a consonant. Given Data / Assumptions: - Word: DETAIL. - Letters: D, E, T, A, I, L. - Vowels: E, A, I (3 vowels). - Consonants: D, T, L (3 consonants). - Positions: 6, labelled from 1 to 6. - Odd positions: 1, 3, 5 (three positions). - Even positions: 2, 4, 6 (three positions). - Condition: vowels must occupy only odd positions, so positions 1, 3, 5 are for vowels, and positions 2, 4, 6 are consequently for consonants. Concept / Approach: Because the number of vowels equals the number of odd positions, and the condition forbids vowels in even positions, the placement of vowels and consonants is structurally forced: all vowels must be placed among the three odd positions, and all consonants must be placed among the three even positions. Within these slots, the vowels and consonants can be permuted independently. We count the permutations of vowels in their slots and multiply by the permutations of consonants in theirs. Step-by-Step Solution: Step 1: Identify the vowels: E, A, I and the odd positions: 1, 3, 5.Step 2: Since there are 3 vowels and 3 odd positions, each odd position will contain exactly one vowel.Step 3: The number of ways to arrange the 3 distinct vowels in positions 1, 3 and 5 is 3!.Step 4: Compute 3! = 3 * 2 * 1 = 6.Step 5: For the even positions (2, 4, 6), we must place the 3 consonants D, T, L.Step 6: The number of ways to arrange 3 distinct consonants in these 3 even positions is also 3! = 6.Step 7: Total valid arrangements = (ways to arrange vowels) * (ways to arrange consonants) = 3! * 3! = 6 * 6.Step 8: Compute 6 * 6 = 36. Verification / Alternative check: We can check that there is no alternative pattern: if any vowel were placed in an even position, the condition would be violated. Likewise, leaving any odd position for a consonant would force a vowel into an even position, again violating the rule. Hence there is a one-to-one correspondence between permutations of vowels and permutations of consonants, as captured in 3! * 3!. Recomputing 3! and the final product confirms the total of 36 arrangements. Why Other Options Are Wrong: - 25 and 42 are not products of two 3! terms and do not arise naturally from correct counting; they result from misusing combinations or missing a factor of permutations. - 120 is the full 5! or a related count that ignores the positional restriction, effectively counting all or most permutations without separating vowels and consonants. Common Pitfalls: Some students misinterpret "vowels occupy only odd positions" as merely prohibiting vowels from even positions but allowing odd positions to be optionally occupied by consonants, which is impossible here because the counts of vowels and odd positions match exactly. Another mistake is to treat the task as a 6! problem and then attempt to divide by something, which does not reflect the structural constraints. Correctly recognising that there are two independent permutations—one for vowels and one for consonants—makes the problem straightforward. Final Answer: The letters of the word DETAIL can be arranged in 36 different ways so that all vowels occupy only the odd positions.

Question 7

A five-letter word such as 'ROTOR' reads the same forwards and backwards and is called a palindrome. Using the 26 letters of the English alphabet, what is the maximum possible number of distinct five-letter palindromes that can be formed?

Accepted Answer

Introduction / Context: This problem concerns palindromic words, which read the same from left to right and right to left. A five-letter palindrome has a specific mirrored structure, which greatly restricts how letters can be chosen compared to arbitrary five-letter words. The question asks for the maximum number of five-letter palindromes that can be formed using the 26 letters of the English alphabet, allowing any letters to be repeated, as long as the palindromic structure is obeyed. Given Data / Assumptions: - We have 26 possible letters for each position (the English alphabet). - We are forming five-letter strings. - Each string must satisfy the palindrome property: it reads the same forwards and backwards. - Letters may be repeated; there is no restriction that all letters must be distinct. Concept / Approach: A five-letter palindrome has the form a b c b a when written from left to right. The first and fifth letters must match, the second and fourth letters must match, and the middle letter can be any letter. Thus, although there are five positions, only three of them are independent: the first, second, and third positions. Once these three are chosen, the last two positions are determined automatically by the palindrome requirement. Therefore, the total number of five-letter palindromes equals the number of ways to choose letters for these three independent positions. Step-by-Step Solution: Step 1: Consider the general structure of a five-letter palindrome: positions 1, 2, 3, 4, 5 must form the pattern a b c b a.Step 2: The letter in position 1 can be any of the 26 letters.Step 3: Once position 1 is chosen, position 5 must be the same letter, and thus has no further independent choice.Step 4: The letter in position 2 can also be any of the 26 letters.Step 5: Once position 2 is chosen, position 4 must match it and is therefore determined.Step 6: The letter in position 3 (the middle letter) is completely free and can again be any of the 26 letters.Step 7: Therefore, the total number of five-letter palindromes is 26 * 26 * 26 = 26^3.Step 8: Compute 26^3 = 26 * 26 * 26 = 676 * 26 = 17576. Verification / Alternative check: You can think of the process as choosing a triple (a, b, c) where each of a, b and c is any letter from the 26-letter alphabet. For each such triple, there is exactly one corresponding palindrome a b c b a. This creates a one-to-one correspondence between the set of all triples and the set of all palindromes, so counting triples directly gives the count of palindromes. Since there are 26 choices for each of the three letters, this again yields 26^3 = 17576. Why Other Options Are Wrong: - 17756 is close but not equal to 26^3, and there is no natural combinatorial structure giving exactly this number for five-letter palindromes. - 12657 and 12666 are also not equal to any simple power or product of 26 associated with independent positions in a five-letter palindrome and therefore do not match the correct structural reasoning. Common Pitfalls: Some learners mistakenly think that all five positions are independent and compute 26^5, which counts all five-letter strings, not just palindromes. Others may incorrectly enforce extra conditions, such as no repeated letters, which is not stated in the problem. The key is recognising that palindromes greatly reduce independence among positions and that only the first three letters need to be freely chosen. Final Answer: The maximum possible number of distinct five-letter palindromes that can be formed using the 26 letters of the English alphabet is 17576.

Question 8

Ten distinct letters are to be posted into five different post boxes. Each post box can receive any number of letters (each can hold more than ten letters without any restriction). In how many different ways can all the ten letters be posted into th…

Accepted Answer

Introduction / Context: This problem tests a very common idea from permutations and combinations, namely distributing distinct objects (letters) into distinct groups (post boxes) when each group can hold any number of objects. Such questions appear frequently in aptitude exams to check whether the student can quickly recognise the correct counting model and avoid overcomplicating the scenario with unnecessary conditions. Given Data / Assumptions: There are 10 distinct letters that need to be posted. There are 5 distinct post boxes available. Each letter must go into exactly one post box. Each post box can hold any number of letters; there is no upper limit. Concept / Approach: The key concept is to observe that each letter has 5 independent choices because it can be posted into any of the 5 post boxes. When choices for different letters are independent, the total number of ways is obtained by multiplying the number of choices for each letter. Thus, the basic rule used here is the fundamental principle of counting: if one action can be done in m ways and another in n ways independently, then together they can be done in m * n ways. Step-by-Step Solution: Step 1: Consider the first letter. It can be placed in any one of the 5 post boxes, so it has 5 choices.Step 2: The second letter is also free to go into any one of the 5 post boxes, again giving 5 choices, independent of the first letter.Step 3: Similarly, each of the remaining 8 letters has 5 choices, because there is no capacity restriction.Step 4: By the fundamental principle of counting, the total number of ways is 5 * 5 * 5 * ... (10 times) which is 5^10.Step 5: Therefore, the required number of ways to post 10 letters into 5 post boxes is 5^10. Verification / Alternative check: An alternative way to think of the same result is to treat each letter as a position in a 10 length string, where for each position we choose one of the 5 boxes as its destination. This again clearly gives 5 choices for each of the 10 positions, hence 5^10. This matches our earlier calculation, so the answer is consistent. Why Other Options Are Wrong: 10^5 is incorrect because it would correspond to distributing 5 letters into 10 boxes, which is not the given situation here. 5P5 is just 5! which counts permutations of the boxes, not distributions of letters. 5C5 counts the number of ways to choose 5 boxes out of 5, which is 1 and is unrelated to the actual question. "None of these" is not correct because one of the listed expressions, namely 5^10, is correct. Common Pitfalls: Students sometimes confuse this model with permutations like 10! or combinations like 10C5, both of which would be wrong because we are not arranging letters in order; we are allocating letters into boxes. Another pitfall is to assume there must be a maximum of one letter or some fixed number of letters per box, which is explicitly not the case here. Correctly recognising that each letter independently chooses a box is essential. Final Answer: The total number of ways to post the 10 letters into the 5 post boxes is 5^10.

Question 9

Five distinct letters k, l, m, n and o are available. Using any three of these letters at a time, and not repeating any letter within a word, how many different three letter words (with or without meaning) can be formed?

Accepted Answer

Introduction / Context: This question checks understanding of permutations when we form words or arrangements from a subset of available distinct letters. The focus is on recognising that order matters in forming three letter words and that each letter can be used at most once in any particular word. This is a standard aptitude topic under permutations and combinations and often appears in competitive exams. Given Data / Assumptions: Available letters are k, l, m, n and o, so there are 5 distinct letters. We need to form words of exactly three letters. No letter is repeated within a word. The words need not have any dictionary meaning; they are simply arrangements. Concept / Approach: Since the words are three letter arrangements where order matters and repetition is not allowed, we use permutations. Specifically, we are finding the number of permutations of 5 distinct letters taken 3 at a time. This is denoted by 5P3, and the formula for nPr is n! / (n - r)!. We will calculate the total by multiplying the number of choices for each position in the word step by step. Step-by-Step Solution: Step 1: For the first letter of the word, we can choose any of the 5 letters, so there are 5 choices.Step 2: For the second letter, one letter has already been used, so 4 letters remain, giving 4 choices.Step 3: For the third letter, two letters are already used, so 3 letters remain, giving 3 choices.Step 4: Multiply the choices: total number of words = 5 * 4 * 3 = 60.Step 5: Therefore, the required number of three letter words is 60. Verification / Alternative check: The result can also be confirmed using the permutation formula 5P3 = 5! / (5 - 3)! = 5! / 2! = (5 * 4 * 3 * 2 * 1) / (2 * 1) = 5 * 4 * 3 = 60. Both the direct counting method and the formula give the same value, so the answer is consistent. Why Other Options Are Wrong: 120 would be 5! and would correspond to arranging all five letters, not just three at a time. 240 is an overestimate and does not correspond to any standard arrangement count in this scenario. 30 would come from 5C3 and would count combinations where order does not matter, which is not appropriate because different orders like klm and mlk are distinct words. Common Pitfalls: A common mistake is to use combinations instead of permutations, thereby forgetting that different orders of letters give different words. Another mistake is to accidentally allow repetition, which would change the calculation to 5^3. Careful reading of the condition "without repetition of alphabets" avoids such errors. Final Answer: The number of three letter words that can be formed is 60.

Question 10

If 1 × 2 × 3 × 4 × ... × n is denoted by n! (n factorial), then simplify the expression 15! - 14! - 13! and express it in factored form.

Accepted Answer

Introduction / Context: This question tests the understanding of factorial notation and algebraic manipulation of expressions involving factorials. Factorial identities appear frequently in combinatorics, probability and series expansions, so being comfortable factoring expressions like 15! - 14! - 13! is important for simplifying more complex formulas in aptitude as well as in higher mathematics. Given Data / Assumptions: n! denotes the product 1 × 2 × 3 × ... × n. We are given the expression 15! - 14! - 13!. The task is to simplify the expression and match it with one of the given factored forms. No approximations are involved; it is pure algebraic simplification. Concept / Approach: The main idea is to factor out the smallest factorial term that is common to all parts of the expression. Since 13! divides both 14! and 15!, it is natural to factor 13! out of the entire expression. After writing 14! and 15! as multiples of 13!, we can combine the coefficients and factor the result further, if possible. This reduces a large looking expression into a compact product form. Step-by-Step Solution: Step 1: Write 15! in terms of 13!: 15! = 15 × 14 × 13!.Step 2: Write 14! in terms of 13!: 14! = 14 × 13!.Step 3: The expression becomes 15! - 14! - 13! = 15 × 14 × 13! - 14 × 13! - 13!.Step 4: Factor out the common 13!: 13! × (15 × 14 - 14 - 1).Step 5: Simplify inside the brackets: 15 × 14 = 210, then 210 - 14 - 1 = 210 - 15 = 195.Step 6: Write 195 as 15 × 13, so the expression becomes 13! × 15 × 13 = 15 × 13 × 13!. Verification / Alternative check: As a numerical check, one could compute the approximate values using a calculator or software to verify that 15! - 14! - 13! equals 15 × 13 × 13!. However, the algebraic derivation above is already exact. The factoring step using 13! is standard and reliable, so the expression 15 × 13 × 13! is confirmed as correct. Why Other Options Are Wrong: 14 × 13 × 13! would correspond to 13! × 182, but we have 13! × 195, so it is too small. 15 × 14 × 14! is much larger and corresponds to 15 × 14 × 14!, which is unrelated. 14 × 12 × 12! is not algebraically equivalent either. Hence only 15 × 13 × 13! matches the simplification. Common Pitfalls: One common error is to factor out 15! or 14! incorrectly, leading to complicated or incorrect expressions. Another mistake is forgetting that 14! = 14 × 13! and 15! = 15 × 14 × 13!, which makes the algebra messy. Always factor out the smallest common factorial term to keep calculations neat. Final Answer: The simplified expression 15! - 14! - 13! equals 15 × 13 × 13!.

Question 11

A box contains 3 blue marbles, 4 red marbles, 6 green marbles and 2 yellow marbles. If two marbles are picked at random without replacement, what is the probability that both marbles picked are from the set of blue or yellow marbles (that is, each s…

Accepted Answer

Introduction / Context: This problem checks basic probability with combinations, where we draw two objects from a group containing several categories. The goal is to find the probability that both drawn marbles come only from a specified subset of colours, namely blue and yellow. Such questions train students to correctly count favourable outcomes and total outcomes using combinations instead of confusing them with ordered selections. Given Data / Assumptions: Blue marbles: 3. Red marbles: 4. Green marbles: 6. Yellow marbles: 2. Two marbles are drawn at random without replacement. We want both marbles to be either blue or yellow, with any combination among them allowed. Concept / Approach: The standard approach is to use combinations to count selections where order does not matter. The total number of ways to choose 2 marbles from all available marbles is computed using nC2. Then we count the favourable ways where both marbles are drawn only from the combined group of blue and yellow marbles. The probability is favourable combinations divided by total combinations. Step-by-Step Solution: Step 1: Count the total number of marbles: 3 + 4 + 6 + 2 = 15 marbles.Step 2: Total ways to choose 2 marbles from 15 marbles is 15C2 = (15 × 14) / 2 = 105.Step 3: The favourable colours are blue and yellow. Number of blue or yellow marbles combined is 3 + 2 = 5.Step 4: Number of ways to choose 2 marbles both from this group of 5 is 5C2 = (5 × 4) / 2 = 10.Step 5: Probability that both marbles are either blue or yellow is favourable / total = 10 / 105.Step 6: Simplify 10 / 105 by dividing numerator and denominator by 5, giving 2 / 21. Verification / Alternative check: We can also check the result by considering the probability step by step. First pick any marble that is blue or yellow: probability = 5 / 15. Then, given that, there are 4 remaining blue or yellow marbles out of 14 total marbles, so the second probability is 4 / 14. Multiply: (5 / 15) * (4 / 14) = (5 * 4) / (15 * 14) = 20 / 210 = 2 / 21. This matches the earlier result, confirming correctness. Why Other Options Are Wrong: 3/17, 4/21 and 5/17 do not match the correct calculation and arise from incorrect counting or incorrect reduction of fractions. For example, 4/21 might come from mistaken counting of favourable cases. None of these values matches the properly derived probability 2/21. Therefore, option 2/21 is the only correct choice. Common Pitfalls: Students often mistakenly use permutations (which consider order) instead of combinations, or they fail to combine the blue and yellow marbles before counting. Another common error is to add probabilities instead of counting combinations. Always decide whether the order of selection matters and use combinations when order is irrelevant in selection problems like this one. Final Answer: The required probability that both marbles are either blue or yellow is 2/21.

Question 12

There are 11 True or False type questions in a test. Each question can be answered independently as either True or False. In how many different ways can a student answer all the 11 questions?

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Introduction / Context: This is a basic question from the topic of counting and binary choices. Each True or False question represents a binary decision, and the total number of different answer patterns for the whole test can be calculated using powers of 2. Such problems help develop comfort with the idea that each independent yes or no type choice doubles the number of possible outcomes. Given Data / Assumptions: There are 11 questions in total. Each question can be answered in exactly two ways: True or False. Answers to different questions are independent of one another. We need the number of complete answer sheets possible for all 11 questions together. Concept / Approach: Because each question has 2 possible responses and the choices across questions are independent, we use the rule for the product of independent choices. If there are n independent binary choices, the total number of possible outcome patterns is 2^n. Here, n is 11. We do not care about correctness of the answers, only distinct answer combinations. Step-by-Step Solution: Step 1: For question 1, there are 2 choices: True or False.Step 2: For question 2, there are again 2 choices, independent of question 1.Step 3: This pattern continues for all 11 questions, each contributing a factor of 2.Step 4: Therefore, the total number of different answer patterns is 2 × 2 × ... (11 times) = 2^11.Step 5: Compute 2^10 = 1024 and then multiply by 2 to get 2^11 = 2048. Verification / Alternative check: Another way to see this is to write the answer pattern as a string of length 11, where each position can be T or F. For each of the 11 positions we have 2 possibilities, and so the total number of distinct strings is 2^11, exactly the same reasoning as above. Calculating 2^11 gives 2048, confirming our result. Why Other Options Are Wrong: 11!/2 and 11! count permutations of 11 distinct objects, which is not relevant here because questions are fixed and answers are only two types. 1024 corresponds to 2^10, which would be correct if there were only 10 questions. Therefore 2048 is the only correct choice among the options. Common Pitfalls: Students sometimes confuse factorial with powers and mistakenly think that the number of ways to answer n questions is n! rather than 2^n. Another common error is miscomputing 2^11, for example stopping at 1024 and forgetting to multiply by 2 one more time. Always remember that each independent binary choice multiplies the total by 2. Final Answer: The number of different ways to answer the 11 True or False questions is 2048.

Question 13

Each vowel in the word NICELY is changed to the previous letter in the English alphabet and each consonant is changed to the next letter in the English alphabet. After these changes, all the resulting letters are arranged in alphabetical order. Whic…

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Introduction / Context: This question combines simple letter coding with alphabetical ordering. It tests the ability to carefully apply transformation rules to each character of a word and then to sort the resulting letters correctly. Such problems are common in reasoning sections of aptitude tests and help verify careful reading and attention to detail. Given Data / Assumptions: Original word: NICELY. Vowels in the word (I and E) are changed to the previous letter in the English alphabet. Consonants in the word (N, C, L and Y) are changed to the next letter in the English alphabet. After all substitutions, the new letters are arranged in strict alphabetical order from left to right. We must identify the fourth letter from the left in this ordered list. Concept / Approach: The problem proceeds in two stages. First, perform a letter by letter transformation using the given coding rule. Second, once the new letters are obtained, arrange them alphabetically and then simply count the letters from left to right. Care must be taken not to confuse the directions of the changes for vowels and consonants. Step-by-Step Solution: Step 1: Identify vowels and consonants in NICELY: N, C, L and Y are consonants; I and E are vowels.Step 2: Apply the rule to each letter one by one:N (consonant) becomes O, the next letter in the alphabet.I (vowel) becomes H, the previous letter in the alphabet.C (consonant) becomes D, the next letter in the alphabet.E (vowel) becomes D, the previous letter in the alphabet.L (consonant) becomes M, the next letter.Y (consonant) becomes Z, the next letter.Step 3: The resulting multiset of letters is {O, H, D, D, M, Z}.Step 4: Arrange these letters alphabetically: D, D, H, M, O, Z.Step 5: Count from the left: first D, second D, third H, fourth M. So the fourth letter from the left is M. Verification / Alternative check: We can quickly recheck by listing the letters again in order and verifying that the transformations are correctly applied. A careful recount of positions in the sorted sequence confirms that after D, D and H, the next distinct letter is M in the fourth position. No other letter order yields a different result, so this is consistent. Why Other Options Are Wrong: J and B do not appear in the transformed set of letters at all. O and Z do appear but they are the fifth and sixth letters when sorted alphabetically, not the fourth. Therefore, among the options given, M is the only correct answer. Common Pitfalls: Students often invert the rule, for example moving consonants to the previous and vowels to the next letter, which leads to an incorrect set of letters and hence a wrong answer. Another common error is to forget that Y is a consonant in this context and treat it incorrectly. Finally, mistakes can occur when sorting the letters alphabetically if the set is not written out clearly. Final Answer: The fourth letter from the left in the final alphabetical arrangement is M.

Question 14

In a bag, the ratio of the number of blue balls to the number of red balls is constant. When there were 44 red balls in the bag, the number of blue balls was 36. If later the number of blue balls becomes 54 while keeping the same ratio, how many red…

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Introduction / Context: This question checks understanding of ratios and proportional reasoning. The ratio between two types of objects is kept constant, and we use the initial information to find the new quantity of one type when the quantity of the other type changes. This is a common type of question in aptitude tests and forms a basic application of ratios and proportions. Given Data / Assumptions: The ratio of blue balls to red balls remains constant at all times. When there are 44 red balls, there are 36 blue balls. Later, the number of blue balls becomes 54. We need to find the corresponding number of red balls when the ratio is unchanged. Concept / Approach: When the ratio between two quantities is constant, they are directly proportional. This means blue / red = constant. We can first compute the ratio using the initial data, then set up a proportion involving the new number of blue balls and the unknown number of red balls. Solving the resulting proportion yields the required value. Step-by-Step Solution: Step 1: Compute the initial ratio of blue to red balls using 36 blue and 44 red.Step 2: Simplify the ratio 36 : 44 by dividing both numbers by their greatest common divisor, which is 4, giving 9 : 11.Step 3: Let the new number of red balls be R when the number of blue balls is 54. Since the ratio stays 9 : 11, we have 54 : R = 9 : 11.Step 4: Use the property of proportions: 54 / R = 9 / 11, so R = 54 × 11 / 9.Step 5: Calculate R: first 54 / 9 = 6, then 6 × 11 = 66.Step 6: Therefore, the number of red balls needed to maintain the same ratio is 66. Verification / Alternative check: We can verify by checking the ratio again with the new numbers: 54 blue and 66 red. The simplified ratio is 54 : 66. Divide both by 6 to get 9 : 11, which is exactly the original ratio. Hence the solution is consistent with the requirement that the ratio remains constant. Why Other Options Are Wrong: Values such as 64, 68, 32 or 72 do not give the ratio 9 : 11 when paired with 54. For example, 54 : 64 simplifies approximately to 27 : 32, which is not the same as 9 : 11. Therefore these options violate the condition of a constant ratio and must be rejected. Common Pitfalls: Many students mistakenly set up the ratio in the reverse order or misapply cross multiplication. Some also forget to simplify the ratio first and try to guess the answer. Systematically writing the proportion and solving it avoids such errors and ensures accuracy in similar ratio based questions. Final Answer: The number of red balls in the bag when there are 54 blue balls is 66.

Question 15

Using all the letters of the word CREATIVITY, how many distinct words or arrangements can be formed, considering that some letters are repeated and that the words may or may not have meaning?

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Introduction / Context: This question tests permutations with repeated letters. When forming arrangements from letters of a word that contains repeated characters, we must adjust the standard formula for permutations to avoid overcounting identical arrangements. This is a standard pattern in aptitude tests for permutations and combinations involving words like COMMITTEE or CREATIVITY. Given Data / Assumptions: The word is CREATIVITY. The letters in order are C, R, E, A, T, I, V, I, T, Y. Total number of letters is 10. There are repeated letters: the letter I appears twice and the letter T appears twice. We count distinct arrangements, so arrangements that differ only by swapping identical letters are considered the same. Concept / Approach: For n objects where some objects are repeated, the number of distinct permutations is given by n! divided by the factorials of the frequencies of repeated objects. Here n is 10, and we have two I's and two T's. Hence we will compute 10! / (2! * 2!). This formula ensures that swapping identical letters does not produce artificially different arrangements in our count. Step-by-Step Solution: Step 1: Count the total number of letters: there are 10 letters in CREATIVITY.Step 2: Count the frequency of each letter: I appears twice, T appears twice, and the remaining letters C, R, E, A, V and Y each appear once.Step 3: Use the permutation formula for repeated letters: total arrangements = 10! / (2! * 2!).Step 4: Compute 10! = 3628800.Step 5: Compute 2! * 2! = 2 * 2 = 4.Step 6: Divide 3628800 by 4 to get 907200 distinct arrangements. Verification / Alternative check: If all letters were distinct, there would be 10! = 3628800 arrangements. However, because each pair of identical letters (I and T) leads to overcounting by a factor of 2 for each pair, we divide by 2 for the first pair and again by 2 for the second pair, which is equivalent to dividing by 4. The resulting count 907200 is reasonable and matches the standard formula for permutations with repeats. Why Other Options Are Wrong: The value 362880 corresponds to 9!, not 10! / (2! * 2!). The other numbers like 851250 or 751210 do not correspond to any correct application of factorial division for repeated letters. Only 907200 is consistent with the correct formula for this situation. Common Pitfalls: Students often forget to divide by factorials of repeated letters, which leads to overcounting. Others miscount the number of repeats, for example, missing that both I and T are repeated in CREATIVITY. Carefully listing frequencies before applying the formula helps avoid such mistakes in similar problems. Final Answer: The number of distinct arrangements that can be formed from the letters of CREATIVITY is 907200.

Question 16

A box contains 5 green pearls, 4 yellow pearls and 5 white pearls. Four pearls are drawn at random without replacement. What is the probability that the four pearls drawn are not all of the same colour?

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Introduction / Context: This probability question involves drawing multiple objects from a box containing several colours and asking for the probability that all selected objects are not of the same colour. It is more convenient to compute the complement event, namely that all four pearls are of the same colour, and subtract this from 1. This complement method is very useful when the direct event has many complicated cases. Given Data / Assumptions: Green pearls: 5. Yellow pearls: 4. White pearls: 5. Total pearls: 5 + 4 + 5 = 14. Four pearls are drawn at random without replacement. We want the probability that the four pearls drawn are not all the same colour. Concept / Approach: Let us define event A as the event that all four pearls drawn are of the same colour. The required probability is then 1 minus P(A). We compute P(A) by summing probabilities of the disjoint subcases where all four are green, or all four are yellow, or all four are white. Each of these probabilities is found using combinations, because the order of drawing does not matter. Step-by-Step Solution: Step 1: Total ways to choose 4 pearls out of 14 is 14C4.Step 2: Compute 14C4 = (14 × 13 × 12 × 11) / (4 × 3 × 2 × 1) = 1001.Step 3: Ways to choose 4 green pearls: since there are 5 green pearls, this is 5C4 = 5.Step 4: Ways to choose 4 yellow pearls: there are 4 yellow pearls, so 4C4 = 1.Step 5: Ways to choose 4 white pearls: there are 5 white pearls, so again 5C4 = 5.Step 6: Total favourable ways for event A (all same colour) = 5 + 1 + 5 = 11.Step 7: Thus, P(A) = 11 / 1001 = 1 / 91.Step 8: Probability that the four pearls are not all of the same colour is 1 - 1 / 91 = 90 / 91. Verification / Alternative check: The computed probability 90 / 91 is very close to 1, which makes intuitive sense because it is quite rare to randomly pick four pearls all of the same colour given the distribution in the box. There are only 11 combinations out of 1001 where all four are of the same colour, confirming that the complement event is highly likely. This supports the correctness of the calculation. Why Other Options Are Wrong: The provided options in the original statement map as follows: Option A = 11/91, Option B = 4/11, Option C = 1/11 and Option D = 90/91. We have just shown that the correct probability is 90/91, so only Option D matches. The others represent either the complement probability or unrelated fractions and therefore are incorrect for the required event. Common Pitfalls: A frequent mistake is to misinterpret the phrase "not of the same colour" and attempt to compute directly the probability that at least two colours appear, which leads to complicated casework. Another mistake is to compute probabilities for only one colour and forget to sum over all colours. Using the complement method simplifies the logic and reduces errors. Final Answer: The probability that the four pearls drawn are not all of the same colour is 90/91, which corresponds to Option D.

Question 17

From the digits 2, 3, 5, 6, 7 and 9, how many different three digit numbers can be formed that are divisible by 5, if no digit is repeated within a number?

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Introduction / Context: This is a standard permutations question with an additional divisibility condition. We have to form three digit numbers from a given list of digits, with no repetition allowed, and the numbers must be divisible by 5. Divisibility by 5 gives a direct condition on the units digit, which significantly simplifies the counting process. Given Data / Assumptions: Available digits are 2, 3, 5, 6, 7 and 9. We must form three digit numbers. No digit is repeated within a number. The number must be divisible by 5. Concept / Approach: A number is divisible by 5 if and only if its units (ones) digit is 0 or 5. Because the digit 0 is not in the given set, the units digit must be 5. Once the units digit is fixed as 5, we choose the hundreds and tens digits from the remaining five digits. Since order of these two positions matters and repetition is not allowed, we use permutations for the remaining positions. Step-by-Step Solution: Step 1: Fix the units digit as 5 to satisfy the condition of divisibility by 5.Step 2: The remaining digits available for the hundreds and tens places are {2, 3, 6, 7, 9}, a total of 5 digits.Step 3: For the hundreds place, we can choose any one of these 5 digits (none of them is zero, so leading zeros are not an issue).Step 4: After fixing the hundreds digit, there are 4 remaining digits to choose from for the tens place.Step 5: Therefore, the total number of three digit numbers that can be formed is 5 × 4 = 20. Verification / Alternative check: We can manually reason with permutations: the number of ways to arrange any two of the five available digits in order is 5P2 = 5 × 4 = 20, and each such pair, when followed by the fixed digit 5, forms a unique three digit number divisible by 5. This confirms the earlier count of 20. Why Other Options Are Wrong: The counts 5, 10, 15 or 25 do not correspond to correct permutations of the digits under the divisibility constraint. For example, 5 would only correspond to fixing the first digit and ignoring the second entirely, which is incorrect. The only count that correctly reflects the choices for both hundreds and tens positions is 20. Common Pitfalls: Students sometimes forget the rule for divisibility by 5 and try to consider many units digits. Others miscount by allowing repetition of digits even though the question explicitly restricts it. A clear stepwise approach, first fixing the units digit and then arranging the remaining digits, helps avoid these errors. Final Answer: The number of three digit numbers divisible by 5 that can be formed is 20.

Question 18

Out of a collection of 7 consonants and 4 vowels, in how many distinct ways can we form a word consisting of exactly 3 consonants and 2 vowels, assuming that order of letters in the word matters?

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Introduction / Context: This problem tests the ability to combine combinations and permutations in one calculation. We first choose which letters will be used and then arrange them in all possible orders. These multi step counting problems are very common in competitive exams and are an essential part of permutations and combinations practice. Given Data / Assumptions: There are 7 distinct consonants. There are 4 distinct vowels. We need to form a word of length 5 with exactly 3 consonants and 2 vowels. No letter is repeated within a single word. Order of letters in the word matters, so different arrangements of the same letters count as different words. Concept / Approach: The process naturally splits into two stages. First, select which 3 consonants from the 7 and which 2 vowels from the 4 will appear in the word. This selection uses combinations. Second, arrange the chosen 5 letters in all possible orders, which uses permutations. The total number of words is the product of the number of ways of selecting letters and the number of ways of arranging them. Step-by-Step Solution: Step 1: Choose 3 consonants out of 7. This can be done in 7C3 ways.Step 2: Compute 7C3 = (7 × 6 × 5) / (3 × 2 × 1) = 35.Step 3: Choose 2 vowels out of 4. This can be done in 4C2 ways.Step 4: Compute 4C2 = (4 × 3) / (2 × 1) = 6.Step 5: After selection, we have 3 + 2 = 5 distinct letters. The number of ways to arrange 5 distinct letters in order is 5! = 120.Step 6: Multiply all factors: total words = 7C3 × 4C2 × 5! = 35 × 6 × 120.Step 7: First multiply 35 × 6 = 210, then multiply 210 × 120 = 25200. Verification / Alternative check: A quick sanity check can be done by noting that 25200 is a reasonable large number for forming ordered words from 11 letters under a moderate restriction. If we had used only permutations directly without separating the selection step, we would risk double counting or miscounting. The combination then permutation method is standard and robust. Why Other Options Are Wrong: The other numerical options do not result from any correct combination of 7C3, 4C2 and 5!. For instance, 25000 and 25225 are very close to the correct answer but do not equal 35 × 6 × 120. They are likely included as traps for arithmetic mistakes. Only 25200 exactly matches the required calculation. Common Pitfalls: Common errors include forgetting to multiply by 5! after selecting the letters, or choosing the wrong combination counts such as 7P3 instead of 7C3. Another mistake is attempting to place consonants and vowels in specific positions prematurely, which is unnecessary here. Following the clear two stage plan avoids such mistakes. Final Answer: The number of different words that can be formed is 25200.

Question 19

A class has 8 football players. A team of 5 members and a separate captain are to be selected from these 8 players (so that there are 6 distinct people: 1 captain and 5 other team members). In how many different ways can such a selection be made?

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Introduction / Context: This selection problem tests the understanding of combinations under an additional role constraint. We not only choose members for a team but also designate one distinct person as captain. It is important to correctly interpret that the captain is not an extra role for a chosen member of the team but a separate person in addition to the five other team members. Given Data / Assumptions: There are 8 distinct football players. A team of 5 members is to be chosen. A separate captain is also to be chosen from the remaining players, making a total of 6 distinct people (5 members plus 1 captain). Order among the 5 team members does not matter, but the captain is a special designated role. Concept / Approach: We should choose the captain first or the team first; both methods yield the same count. The most straightforward approach is to first choose the captain from all 8 players, and then choose the remaining 5 team members from the remaining 7 players. Since the captain is not counted among the 5, the selection steps are clearly separated. The total number of ways is the product of the ways of these two choices. Step-by-Step Solution: Step 1: Choose the captain from the 8 players. There are 8 possible choices for the captain.Step 2: After selecting the captain, there are 7 players left.Step 3: Choose 5 team members from these remaining 7 players. This can be done in 7C5 ways.Step 4: Compute 7C5 = 7C2 = (7 × 6) / (2 × 1) = 21.Step 5: Total number of ways = 8 (choices for captain) × 21 (choices for team) = 168. Verification / Alternative check: An alternative perspective is to first choose the group of 6 people who will be involved (5 members plus the eventual captain). This can be done in 8C6 = 28 ways. From each such selected group of 6, we can then choose 1 of the 6 to be the captain in 6 ways, and the remaining 5 automatically become the team members. Thus total ways would be 8C6 × 6 = 28 × 6 = 168, which matches our previous result. Why Other Options Are Wrong: 210 would be 7C5 × 6 if misapplied, or 8C4 × something, and does not correctly describe the two stage selection. 1260 and 10!/6! are much larger and not connected to the correct combinatorial structure here. Only 168 correctly counts the combinations consistent with the problem statement. Common Pitfalls: The biggest source of error is misinterpreting whether the captain is one of the five team members or an additional separate person. If students assume the captain is chosen from within the team, they will compute 8C5 × 5, which equals 56 × 5 = 280, a different answer. Carefully reading that the team of 5 and the captain are separate roles is essential. Final Answer: The number of different selections of a 5 member team and a separate captain is 168.

Question 20

There are 36 identical books that must be arranged in rows so that each row contains the same number of books. Each row must have at least 3 books and there must be at least 3 rows. A row is defined as a line of books parallel to the front of the ro…

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Introduction / Context: This problem is about finding the number of ways to arrange identical objects into equal sized rows subject to minimum constraints on both the number of rows and the number of books in each row. Such questions are directly linked to the concept of factors and divisors of a number, and they frequently appear in basic number theory sections of aptitude tests. Given Data / Assumptions: Total number of identical books: 36. The books are to be arranged in rows, each row having the same number of books. Each row must contain at least 3 books. There must be at least 3 rows. We count two arrangements as different if they have a different number of rows or a different number of books per row. Concept / Approach: Let the number of rows be r and the number of books in each row be k. The conditions imply r × k = 36, with r ≥ 3 and k ≥ 3. Thus we need to find all factor pairs (r, k) of 36 that satisfy these inequalities. Each valid factor pair corresponds to one distinct arrangement pattern, because the books are identical and only the grid structure matters. Step-by-Step Solution: Step 1: List all factor pairs of 36: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (9, 4), (12, 3), (18, 2), (36, 1).Step 2: Apply the condition r ≥ 3 and k ≥ 3. This removes pairs where rows or books per row are less than 3.Step 3: Discard (1, 36), (2, 18), (18, 2) and (36, 1) because at least one of the numbers is less than 3.Step 4: The valid pairs that remain are (3, 12), (4, 9), (6, 6), (9, 4) and (12, 3).Step 5: Each of these five pairs corresponds to a distinct arrangement: 3 rows of 12, 4 rows of 9, 6 rows of 6, 9 rows of 4 and 12 rows of 3 books.Step 6: Therefore, the number of different arrangements that satisfy the conditions is 5. Verification / Alternative check: We can verify by explicitly checking the constraints for each valid pair. For (3, 12), we have 3 rows and 12 books per row, both at least 3. The same holds for (4, 9), (6, 6), (9, 4) and (12, 3). No other factor pair meets both minimum requirements, so we have found all possibilities. Hence the count 5 is confirmed. Why Other Options Are Wrong: Options 6, 7 or 8 would require more valid factor pairs than actually exist. Since 36 has only the listed factor pairs, and only 5 of them satisfy r ≥ 3 and k ≥ 3, any number greater than 5 must be incorrect. Option 4 is too small and would imply that one of the valid factor pairs had been missed or incorrectly eliminated. Common Pitfalls: Students sometimes forget that a pair like (3, 12) is distinct from (12, 3) only if the definition of rows versus columns is fixed. In this problem, the number of rows is the first coordinate, so both pairs are considered distinct valid arrangements. Another common error is to ignore the minimum conditions and count all factors without filtering, which would lead to 9 instead of 5. Final Answer: The number of different valid arrangements is 5.

Permutation and Combination Questions