Question 1

A committee of 4 people is to be formed from a group of 9 distinct people. If there are no additional restrictions, in how many different ways can this 4-person committee be chosen?

Accepted Answer

Introduction / Context: This is a straightforward combinations question about forming a committee. We want to choose a subset of people from a larger group, and the order in which they are chosen does not matter. Problems of this kind occur frequently in selection scenarios like committee formation, project teams, or representative groups. Given Data / Assumptions: Total people available: 9 distinct individuals. Number of people needed in the committee: 4. There are no special roles like president or secretary; all committee members are equal. Order of selection does not matter; a set of 4 people is counted once. No person can be selected more than once. Concept / Approach: When selecting r people from n distinct individuals and only the set of chosen people matters, we use combinations. The formula is: nCr = n! / (r! * (n - r)!). Here, n = 9 and r = 4. So we compute 9C4 to find the total number of possible 4 person committees. Step-by-Step Solution: Step 1: Identify that n = 9 and r = 4. Step 2: Apply the combination formula: 9C4 = 9! / (4! * 5!). Step 3: Express 9! / 5! as 9 * 8 * 7 * 6. Step 4: Compute the numerator: 9 * 8 * 7 * 6 = 3024. Step 5: Denominator is 4! = 4 * 3 * 2 * 1 = 24. Step 6: Divide 3024 by 24 to get 126. Step 7: Therefore, there are 126 distinct 4 person committees. Verification / Alternative check: We can also compute 9C4 using symmetry: 9C4 = 9C5 because choosing 4 members to include is equivalent to choosing 5 members to leave out. Calculating 9C5 = 9! / (5! * 4!) leads to the same result, 126. This symmetry gives a quick cross check that our answer is consistent. Why Other Options Are Wrong: 120: Close to 5!, but that belongs to a different counting context and not to 9C4. 162 and 170: Both values do not arise from the combination formula with n = 9 and r = 4 and indicate arithmetic or conceptual mistakes. Only 126 fits the correct combination computation. Common Pitfalls: A frequent error is to use permutations instead of combinations, for example computing 9P4 instead of 9C4, which overcounts each committee many times due to different orders. Another mistake is failing to simplify the factorial expression correctly, especially when cancelling numerator and denominator factors. Writing the computation step by step and remembering that committee members have no order prevents these issues. Final Answer: The number of different 4 person committees that can be formed from 9 people is 126.

Question 2

A pizza shop offers 7 different toppings. If a special pizza must have exactly 3 distinct toppings chosen from these 7 and the order of toppings on the pizza does not matter, how many different 3-topping pizzas can be made?

Accepted Answer

Introduction / Context: This problem involves combinations in a simple and familiar context: choosing pizza toppings. We are selecting topping combinations where placement order on the pizza does not matter, only which toppings are included. This is a very typical example used to illustrate combinations in many introductory probability and combinatorics courses. Given Data / Assumptions: Total types of toppings available: 7. Each topping is distinct. We must choose exactly 3 different toppings for a special pizza. The same topping is not chosen more than once. The order in which toppings are chosen or appear on the pizza does not matter. Concept / Approach: Since we are selecting a set of toppings and only the combination matters, this is a pure combination question. The number of ways to choose r toppings from n available toppings is given by: nCr = n! / (r! * (n - r)!). Here, n = 7 and r = 3, so we must compute 7C3 to get the number of different 3 topping pizzas. Step-by-Step Solution: Step 1: Identify n = 7 and r = 3. Step 2: Use the combination formula: 7C3 = 7! / (3! * 4!). Step 3: Simplify 7! / 4! to 7 * 6 * 5. Step 4: Compute numerator: 7 * 6 * 5 = 210. Step 5: Denominator is 3! = 3 * 2 * 1 = 6. Step 6: Divide 210 by 6 to get 35. Step 7: Therefore, there are 35 distinct 3 topping pizza combinations. Verification / Alternative check: We can also reason that once you pick any 3 toppings, there is exactly one pizza corresponding to that set, since order and layout are not considered different. Checking with symmetry, 7C3 equals 7C4 as well, because choosing 3 toppings to include is equivalent to choosing 4 to exclude. Computing 7C4 leads to the same result, confirming 35 as correct. Why Other Options Are Wrong: 49: Equal to 7^2, which would correspond to ordered pairs, not unordered triples. 27: Equal to 3^3, unrelated to choosing 3 items from 7. 25: A random smaller number that does not come from the combination formula. Only 35 correctly represents 7C3. Common Pitfalls: Some students mistakenly think that order matters and use permutations, leading to a much larger number than needed. Others might misinterpret the question and allow repeated toppings, which is not intended when we say distinct toppings. Remember that for combinations such as toppings, flavors, or colors, unless explicitly ordered, we use nCr, not nPr. Final Answer: The number of different 3 topping pizzas that can be made from 7 toppings is 35.

Question 3

You have a collection of 38 distinct songs and you want to create a mix CD that contains exactly 17 different songs. If the order of songs on the CD does not matter (only which songs are included matters), how many different 17-song CDs can be made?

Accepted Answer

Introduction / Context: This question is a large scale combinations problem. We are selecting a subset of songs from a larger collection to make a mix CD, and the order of the songs is not important for counting distinct CDs. For such problems, combinations are used, and the numbers involved can become very large. Understanding how to express and compute these combinations is useful in probability, statistics, and combinatorics. Given Data / Assumptions: Total number of distinct songs available: 38. We want to include exactly 17 different songs on the CD. No song is repeated on the CD. Order of songs on the CD is not considered; only the set of songs matters. Every choice of 17 distinct songs corresponds to one possible mix CD. Concept / Approach: This is a direct combinations problem. We must choose 17 songs from 38 without regard to order. The number of such selections is given by: nCr = n! / (r! * (n - r)!). Here, n = 38 and r = 17, so the required count is 38C17. This is a large number, but we can still express or compute it using the formula or using symmetry 38C17 = 38C21. Step-by-Step Solution: Step 1: Recognize that the answer is 38C17. Step 2: Write 38C17 = 38! / (17! * 21!). Step 3: In practice, this is computed by carefully simplifying the factorial ratio, often using software or a scientific calculator. Step 4: The exact computed value is 28781143380. Step 5: Therefore, there are 28781143380 distinct choices of 17 songs out of 38. Step 6: Each such choice corresponds to a unique unordered mix CD. Verification / Alternative check: We can check the reasonableness using symmetry: 38C17 = 38C21. Since the central binomial coefficients grow very large, it is normal that the number is on the order of tens of billions. A calculator that supports nCr will confirm that 38C17 equals 28781143380. Any significantly smaller number would underestimate the true count of combinations. Why Other Options Are Wrong: 2878114338: This is exactly ten times smaller than the true value and could result from missing a factor of 10 in computation. 29000000000 and 28000000000: These are rough rounded figures that do not match the exact combinatorial calculation of 38C17. Only 28781143380 matches the correct combination value. Common Pitfalls: One common mistake is to treat the order of songs as important and use permutations, which gives a massive overcount. Another is to underestimate the size of binomial coefficients and think the answer must be a much smaller number such as thousands or millions. Some students also incorrectly use 38^17, which is not a combination and allows repetition and ordering. Always pay attention to whether order matters and whether repetition is allowed. Here, it is a pure selection problem without repetition, so 38C17 is the correct model. Final Answer: The number of different unordered 17 song CDs that can be made from 38 distinct songs is 28781143380.

Question 4

A school committee of 5 students is to be formed from a class of 12 students. If John must be on the committee, in how many different committees can be formed under this condition?

Accepted Answer

Introduction / Context: This is a committee selection problem with a simple condition: one particular student, John, must be included. The presence of a mandatory member slightly reduces the problem to choosing the remaining members from the remaining pool. These types of constraints are very common in exam questions related to combinations and selection with conditions. Given Data / Assumptions: Total students in the class: 12. We must form a committee of 5 students. John is one of the 12 students and must be included in every valid committee. All students are distinct individuals. Committees are sets; the order of members does not matter. Concept / Approach: Since John must be in the committee, we fix him as one member. That leaves 4 more positions to be filled. The remaining available students are the other 11 students (everyone except John). Thus the problem reduces to choosing 4 students from these remaining 11 without any further restrictions. That is a combination problem: 11C4. Step-by-Step Solution: Step 1: Reserve one place on the committee for John. Step 2: Remaining positions to fill = 5 - 1 = 4. Step 3: Remaining students available for these 4 positions = 12 - 1 = 11. Step 4: Since order does not matter, the number of ways to choose the remaining 4 members is 11C4. Step 5: 11C4 = 11! / (4! * 7!). Step 6: We can compute this if needed, but the expression 11C4 already correctly represents the count. Step 7: Therefore, the desired number of committees is 11C4. Verification / Alternative check: We can compute 11C4 numerically: 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 792. This confirms that there are 792 valid committees. The symbolic option 11C4 is correctly the combination expression that gives this number, so using it as the answer is appropriate when options are in terms of combinatorial notation. Why Other Options Are Wrong: 11P4: This counts ordered selections, which is not relevant for committees where order does not matter. 11P5: This would correspond to ordered selections of 5 from 11, and it also ignores the fixed presence of John. 11C5: This would mean choosing 5 students from 11 without including John, which does not reflect the requirement that John must be on the committee. Only 11C4 correctly represents choosing the remaining 4 members among 11 when John is already included. Common Pitfalls: Some students forget to account for the fact that John is fixed and still use 12C5, which ignores the condition. Others misinterpret the notation and pick a permutation expression like 11P4, which would treat different orders of the same committee as distinct, overcounting heavily. Always carefully read conditions about required members and whether the order of selection matters. Final Answer: The number of different committees of 5 students that include John is 11C4 (numerically equal to 792).

Question 5

From a standard deck of 52 cards, a 7-card hand is dealt. In how many distinct 7-card hands will there be exactly 2 spades and exactly 3 diamonds (with the remaining 2 cards coming from the other suits)?

Accepted Answer

Introduction / Context: This question is a conditional card combinations problem. We must count 7 card hands with a specified composition of suits: exactly 2 spades, exactly 3 diamonds, and the remaining 2 cards from the other suits (hearts and clubs). These partitioned counting problems are very common in probability and combinatorics involving card games and suit constraints. Given Data / Assumptions: Standard deck: 52 cards, 13 in each suit (spades, hearts, diamonds, clubs). We deal a hand of 7 cards. Exactly 2 cards must be spades. Exactly 3 cards must be diamonds. The remaining 2 cards must come from the other 26 cards (hearts and clubs). Order of cards in the hand does not matter; only the set of cards matters. Concept / Approach: We choose cards suit by suit using combinations: Choose 2 spades out of 13 spades: 13C2. Choose 3 diamonds out of 13 diamonds: 13C3. Choose 2 remaining cards from the 26 non spade, non diamond cards (hearts and clubs): 26C2. Since these choices are independent and define different subsets, we multiply the three combination counts to get the total number of valid 7 card hands. Step-by-Step Solution: Step 1: Count ways to choose spades: 13C2. Step 2: Count ways to choose diamonds: 13C3. Step 3: Cards not spades or diamonds: 52 - 13 - 13 = 26. Step 4: Count ways to choose remaining 2 cards from these 26: 26C2. Step 5: Multiply all three: total hands = 13C2 * 13C3 * 26C2. Step 6: Compute: 13C2 = 78, 13C3 = 286, 26C2 = 325. Step 7: Multiply: 78 * 286 * 325 = 7250100. Verification / Alternative check: We can verify intermediate values: 13C2 = 13 * 12 / 2 = 78, and 13C3 = (13 * 12 * 11) / (3 * 2 * 1) = 286. For 26C2, (26 * 25) / 2 = 325. These are standard combination values. Multiplying 78 * 286 gives 22308, and 22308 * 325 equals 7250100, which matches the given correct option. This confirms that all arithmetic steps are consistent. Why Other Options Are Wrong: 7690030 and 7250000: These are large numbers, but they do not equal the exact product of the correct combination factors. 3454290: Roughly half the true value, suggesting a missing factor or incomplete multiplication. Only 7250100 matches 13C2 * 13C3 * 26C2. Common Pitfalls: One common mistake is to forget that the remaining 2 cards must come from the suits other than spades and diamonds, accidentally counting hands with more than 2 spades or more than 3 diamonds. Another is to treat the 7 cards as ordered and use permutations, which overcounts the number of distinct hands. Carefully partitioning by suit and using combinations for each subset avoids these issues. Final Answer: The number of 7 card hands with exactly 2 spades, exactly 3 diamonds, and the remaining 2 cards from hearts or clubs is 7250100.

Question 6

From a standard deck of 52 cards, a 5-card hand is dealt. If the queen of spades and the four of diamonds must both be included in the hand, how many distinct 5-card hands are possible?

Accepted Answer

Introduction / Context: This card problem asks you to count how many 5 card hands contain two specific cards: the queen of spades and the four of diamonds. Such problems highlight the idea of fixing certain required cards and then choosing the remaining cards freely from the rest of the deck. This is a very common pattern in conditional counting problems in probability and combinatorics. Given Data / Assumptions: Standard deck with 52 distinct cards. We deal a 5 card hand. Two specific cards, the queen of spades and the four of diamonds, must be in every valid hand. The remaining 3 cards can be any cards except those two already required. Order of cards in the hand does not matter; only the set of 5 cards matters. Concept / Approach: We first fix the two required cards in the hand. Once these are chosen, they occupy 2 of the 5 positions. We then need to choose the remaining 3 cards from the deck excluding these two cards. That leaves 52 - 2 = 50 cards available. The number of ways to choose 3 cards from these 50, with order not mattering, is given by the combination 50C3. Step-by-Step Solution: Step 1: Include the queen of spades in the hand. Step 2: Include the four of diamonds in the hand. Step 3: Now 2 of the 5 cards are fixed; we still need 3 more cards. Step 4: Cards remaining in the deck that are eligible: 52 - 2 = 50. Step 5: The number of ways to choose 3 more cards from these 50 is 50C3. Step 6: So the number of distinct 5 card hands containing both fixed cards is 50C3. Verification / Alternative check: If we want a numerical value, compute 50C3: 50! / (3! * 47!) = (50 * 49 * 48) / (3 * 2 * 1) = 19600. Any hand counted this way has the two required cards plus 3 chosen from the remaining 50, and no hand is counted twice. The combination expression 50C3 exactly represents this process, so it is the correct form among the options provided. Why Other Options Are Wrong: 52C5: Counts all possible 5 card hands without any requirement, including many that do not contain both required cards. 52C4: Would correspond to choosing 4 cards from 52, which does not match the structure of our 5 card hand with two fixed cards. 50C4: This would arise if we were choosing 4 remaining cards instead of 3, but that would create a 6 card hand together with the 2 fixed cards. Only 50C3 correctly models choosing the remaining 3 cards from the 50 not fixed cards. Common Pitfalls: Some learners forget to subtract the two fixed cards from the deck and mistakenly use 52C3. Others try to count the ways to place the fixed cards in positions inside the hand, but because a hand is just a set of cards, that positional counting is unnecessary and leads to overcounting. The correct strategy is always to fix required elements and then choose remaining elements from what is left. Final Answer: The number of distinct 5 card hands that contain both the queen of spades and the four of diamonds is 50C3 (numerically equal to 19600).

Question 7

You have 12 different types of candies and you want to choose a selection that contains at least one candy. In how many different ways can you choose one or more candies from these 12 distinct types?

Accepted Answer

Introduction / Context: This problem involves counting subsets of a set of distinct items, in this case 12 different types of candies. We are asked to count all selections that contain at least one candy, meaning empty selections are not allowed. This is a standard application of the relationship between the total number of subsets and the number of non empty subsets. Given Data / Assumptions: There are 12 distinct types of candies. Each type can be either chosen or not chosen in a selection. We want selections that include at least one type of candy. We do not limit the maximum number of types; we can choose anywhere from 1 up to all 12 types. Candies are considered by type; choosing the same type multiple times is not part of this counting. Concept / Approach: For a set of n distinct elements, the total number of subsets (including the empty set) is 2^n. This counts every pattern of choosing or not choosing each element. To find the number of non empty subsets, we subtract the single empty subset from the total. Therefore, for n items, the number of non empty subsets is 2^n - 1. Here, n = 12. Step-by-Step Solution: Step 1: Compute the total number of subsets of a 12 element set: 2^12. Step 2: 2^12 = 4096. Step 3: Among these subsets, exactly one is the empty set, which contains no candies. Step 4: The problem asks for selections with at least one candy, so we must exclude the empty set. Step 5: Therefore, number of non empty subsets = 2^12 - 1. Step 6: 4096 - 1 = 4095. Step 7: Hence, there are 4095 different ways to choose one or more candies. Verification / Alternative check: We can also think of each candy type as a binary choice: either included or not included in the selection. Each of the 12 types contributes a factor of 2 choices. Multiplying gives 2 * 2 * ... * 2 (12 times) = 2^12 = 4096 total patterns. Only one pattern has all types excluded, so removing that pattern leaves 4095 valid non empty selections. This matches the earlier reasoning exactly. Why Other Options Are Wrong: 4054, 4050, 4059: These numbers are all less than 4095 and do not have any special relation to powers of 2 for n = 12. None of them equal 2^12 - 1, which is the correct formula. Common Pitfalls: A common mistake is to forget the empty set and answer 4096 instead of 4095. Others may try to sum combinations for choosing 1, 2, 3, ..., 12 types separately, which is correct but time consuming. If done correctly, that sum should also equal 2^12 - 1. Remembering the subset formula and the simple subtraction of the empty set provides a fast and reliable method. Final Answer: The number of different ways to choose one or more types of candy from 12 distinct types is 4095.

Question 8

How many different three-letter words can be formed using the letters of the word HEXAGON, if no letter is repeated in a word?

Accepted Answer

Introduction / Context: This problem tests basic permutation ideas. We are given a word with several distinct letters and asked to count how many different three-letter arrangements or “words” can be formed if repetition of letters is not allowed. Such questions are very common in aptitude exams and directly use the fundamental principle of counting and the concept of permutations. Given Data / Assumptions: The word HEXAGON has 7 distinct letters: H, E, X, A, G, O, N. We want to form three-letter words using these letters. No letter may be repeated within a single three-letter word. The order of letters matters. For example, HEG and GEH are considered different words. Concept / Approach: Whenever we arrange r distinct objects chosen from n distinct objects, and the order is important, we use permutations. The number of permutations is nPr, which equals n * (n - 1) * (n - 2) ... for r factors. Here n = 7 and r = 3. So the count is 7 * 6 * 5. Step-by-Step Solution: Step 1: Count available letters. HEXAGON contains 7 distinct letters, so n = 7.Step 2: We need three-letter words with no repetition, so r = 3.Step 3: For the first position, we can choose any of the 7 letters.Step 4: After choosing one letter, 6 letters remain for the second position.Step 5: After choosing two letters, 5 letters remain for the third position.Step 6: Total number of words = 7 * 6 * 5 = 210. Verification / Alternative check: We can also use the permutation notation directly: 7P3 = 7 * 6 * 5 = 210. This matches our stepwise reasoning, so the result is consistent. Why Other Options Are Wrong: 120 assumes 5 * 4 * 3, as if only 5 letters were available or repetition rules were misunderstood. 160 is not a product of three consecutive integers from 7 and does not match any correct permutation count here. 200 is close to 210 but still incorrect and often appears as a trap when candidates miscalculate one of the factors. Common Pitfalls: Candidates often confuse permutations with combinations and may attempt to use only 7C3, which would ignore ordering. Others forget that the letters of HEXAGON are all distinct and may wrongly subtract for repeated letters that do not exist. Another mistake is to stop after computing 7 * 6 and forget to multiply by 5 for the third position. Final Answer: The number of different three-letter words that can be formed is 210.

Question 9

In how many different ways can three students be assigned to four chartered accountants, if each chartered accountant can take at most one student?

Accepted Answer

Introduction / Context: This question checks understanding of permutations and the idea of assigning people to distinct positions. We have fewer students than chartered accountants, and each accountant can supervise at most one student. We must count the number of possible assignments where all three students are placed with different accountants. Given Data / Assumptions: There are 3 distinct students. There are 4 distinct chartered accountants. Each accountant can supervise at most one student. Each student must be assigned to exactly one accountant. Concept / Approach: We are assigning 3 distinct students to 4 distinct “positions” (the accountants), with no two students going to the same accountant. This is a permutation or injective mapping from 3 students into 4 accountants. The number of such assignments is 4P3, which can be computed as 4 * 3 * 2. Step-by-Step Solution: Step 1: Choose an accountant for the first student. There are 4 choices.Step 2: After assigning the first student, only 3 accountants remain free for the second student.Step 3: After assigning the second student, only 2 accountants remain free for the third student.Step 4: Total number of assignments = 4 * 3 * 2 = 24. Verification / Alternative check: Using permutation notation, the number of ways of assigning 3 students to 4 accountants is 4P3. We know 4P3 = 4! / (4 - 3)! = 4! / 1! = 24. This matches the direct multiplication method, so the answer is confirmed. Why Other Options Are Wrong: 12 would correspond to 4 * 3, which ignores the arrangements for the third student. 36 does not match any correct permutation formula in this setup and usually comes from random multiplication or adding extra factors incorrectly. 16 may come from thinking that each of the 4 accountants has 2 possible states (assigned or not), which does not model the actual assignment process. Common Pitfalls: A frequent mistake is to think in terms of combinations and compute 4C3, which counts only which accountants are used, not which student is attached to whom. Another error is to assume students are identical and ignore ordering, which is not correct here. We must respect that both students and accountants are distinct individuals. Final Answer: The number of ways to assign the three students is 24.

Question 10

When Dr Ram arrives at his clinic, he finds 12 patients waiting. He can see only one patient at a time and exactly three of the patients leave before being seen. In how many different ways can this happen, considering which three leave and the order…

Accepted Answer

Introduction / Context: This problem combines combinations and permutations. We must decide which patients leave without being seen and in what order the remaining patients are seen by the doctor. Each patient is distinct, and the sequence of consultations matters. This is a classic arrangement problem that often appears in competitive exams. Given Data / Assumptions: There are 12 distinct patients. Exactly 3 patients leave without being seen. The remaining 9 patients are all seen by Dr Ram. Dr Ram sees one patient at a time, so the order of these 9 patients matters. Concept / Approach: We split the task into two independent parts. First, we choose which 3 of the 12 patients leave. Second, we arrange the remaining 9 patients in an ordered sequence for consultation. The first part uses combinations, because leaving patients are not ordered. The second part uses permutations, because consultation order matters. The final answer is the product of these two counts. Step-by-Step Solution: Step 1: Choose 3 patients out of 12 to leave. This can be done in 12C3 ways.Step 2: Compute 12C3 = (12 * 11 * 10) / (3 * 2 * 1) = 220.Step 3: The remaining 9 patients will all be seen, and their order matters, so they can be arranged in 9! ways.Step 4: Compute 9! = 362880.Step 5: Total number of possible situations = 12C3 * 9! = 220 * 362880 = 79833600. Verification / Alternative check: We can think of first arranging all 12 patients and then marking 3 of them as leaving instead of being seen. However, marking positions for leaving among a 12-long sequence while preserving the order of the remaining patients leads to the same multiplication of 12C3 and 9!. This confirms our count without redoing the entire calculation. Why Other Options Are Wrong: 479001600 equals 12! and would be correct only if all patients were seen, which is not the case here. 34879012 and 67800983 are random-looking numbers and do not correspond to any natural product of factorials or combinations in this context. Common Pitfalls: Many candidates forget to separate the choice of who leaves from the ordering of who is seen. Some simply compute 9! or 12! and ignore the leaving condition. Others may incorrectly compute 12P9 instead of 12C3 * 9!, which double counts many arrangements. Always break multi-step counting problems into logical stages and multiply only when stages are independent. Final Answer: The number of different possible ways is 79833600.

Question 11

In how many distinct ways can four people sit around a circular table, where two seatings are considered the same if one can be rotated to obtain the other?

Accepted Answer

Introduction / Context: This question tests circular permutations. When people sit around a round table, rotating everyone together does not create a new arrangement, so we must adjust the usual linear permutation formula. Understanding this adjustment is crucial for many seating arrangement questions in aptitude tests. Given Data / Assumptions: There are 4 distinct people. They are seated around a circular table. Two arrangements are considered identical if one can be rotated to look like the other. Concept / Approach: For n distinct objects arranged in a circle, the number of distinct circular permutations is (n - 1)!. This formula accounts for the fact that rotating everyone together does not change the relative seating. Here n = 4, so we use (4 - 1)! = 3!. Step-by-Step Solution: Step 1: If the seats were in a straight line, the number of permutations would be 4! = 24.Step 2: On a circular table, each circular arrangement can be rotated into 4 different linear arrangements.Step 3: Therefore, we divide the linear count by 4 to remove this overcounting.Step 4: Number of circular arrangements = 4! / 4 = 24 / 4 = 6.Step 5: This is equal to (4 - 1)! = 3! = 6, which confirms the result. Verification / Alternative check: We can fix one person’s position as a reference, because the table is circular. If we imagine one person always sitting at the “top” position, the remaining 3 people can be arranged in 3! = 6 ways around them. This is another way to see that the answer is 6. Why Other Options Are Wrong: 3 undercounts the possible orders and might come from a mistaken belief that only limited swaps are distinct. 9 does not correspond to any standard circular permutation formula for 4 distinct objects. 12 is half of 24 and would be correct only if we identified some, but not all, rotations or reflections, which is not the condition given in the question. Common Pitfalls: A common error is to forget the circular nature and simply answer 4!, which is 24. Another mistake is to divide by 2 instead of by 4, treating rotations and reflections together without clearly defining what counts as the same. Always remember: for circular permutations where only rotation is ignored and objects are distinct, the correct formula is (n - 1)!. Final Answer: The number of distinct circular arrangements is 6.

Question 12

In how many ways can five different toys be packed into three identical boxes so that no box is empty, if each box may contain any number of the toys?

Accepted Answer

Introduction / Context: This question tests the idea of distributing distinct objects into identical boxes with a non-emptiness condition. Because the boxes are identical, we only care about which toys are grouped together, not about the labels or positions of the boxes. That makes this a set-partition problem rather than a simple distribution into labelled boxes. Given Data / Assumptions: There are 5 distinct toys. There are 3 boxes, which are identical (unlabelled). No box is allowed to be empty. Any box can hold any number of toys. Concept / Approach: Since the boxes are identical and none can be empty, we must partition the set of 5 toys into exactly 3 nonempty unlabeled groups. The number of such partitions is given by the Stirling number of the second kind S(5, 3). For a small value like 5, we can compute this directly using known values or a simple counting argument. Step-by-Step Solution: Step 1: Recognise that the boxes are identical, so arrangements that differ only by swapping boxes are the same.Step 2: We need to partition 5 distinct elements into 3 nonempty unlabeled subsets.Step 3: The required count is S(5, 3), the Stirling number of the second kind for n = 5 and k = 3.Step 4: Using the known table or formula, S(5, 3) = 25.Step 5: Therefore, there are 25 distinct ways to pack the toys into three identical nonempty boxes. Verification / Alternative check: We can list distributions by type of group sizes: 3 + 1 + 1 and 2 + 2 + 1, and then count the number of ways to choose which toys form each group, ensuring we do not double count due to identical boxes. Doing this carefully also leads to a total of 25 partitions, which confirms the Stirling number value. Why Other Options Are Wrong: 10 and 15 are too small and arise if we only consider one group size pattern or incorrectly treat boxes as labelled or partly labelled. 20 is closer but still undercounts, often because some patterns of group sizes are omitted or double counting is removed incorrectly. Common Pitfalls: Many students mistakenly treat the boxes as distinct and simply distribute toys into three labelled boxes, then try to adjust for identical boxes in an ad hoc way. Others forget the nonempty condition and include partitions with empty boxes. For identical boxes with a fixed number of nonempty boxes, always think in terms of partitions of a set into unlabeled subsets and, when possible, use Stirling numbers of the second kind. Final Answer: The number of ways to pack the toys is 25.

Question 13

In how many different ways can six distinct balls be distributed into four distinct boxes, if each box can hold any number of balls including zero?

Accepted Answer

Introduction / Context: This is a standard distribution problem where distinct objects are placed into distinct boxes. There is no restriction on how many balls each box may contain, and a box may even remain empty. Such problems are solved by considering the independent choices for each object. Given Data / Assumptions: There are 6 distinct balls. There are 4 distinct boxes. Each box can hold any number of balls, including zero. There is no restriction about nonempty boxes. Concept / Approach: Each ball can independently be placed into any one of the 4 boxes. Because the balls are distinct, every distinct assignment of balls to boxes is a different arrangement. For each ball we have 4 choices, and these choices multiply because they are independent. Therefore, the number of arrangements is 4^6. Step-by-Step Solution: Step 1: Consider the first ball. It has 4 possible boxes.Step 2: The second ball also has 4 choices, regardless of where the first ball went.Step 3: The same reasoning applies to each of the 6 balls.Step 4: By the multiplication principle, the total number of ways = 4 * 4 * 4 * 4 * 4 * 4 = 4^6.Step 5: Compute 4^6 = (4^3)^2 = 64^2 = 4096. Verification / Alternative check: We can think of assigning to each ball a code from 1 to 4 representing the chosen box. There are 4 choices for each of the 6 positions in this code, which again gives 4^6 possible codes. This matches the earlier calculation, confirming 4096 as the correct count. Why Other Options Are Wrong: 2048 equals 2^11 and has no direct connection to the 4^6 structure here. It might come from halving 4096 without justification. 1296 equals 6^4 and may be chosen by mistakenly swapping the roles of balls and boxes and using 6^4 instead of 4^6. 576 is 24^2 and does not come from any correct counting method for this scenario. Common Pitfalls: A common mistake is to treat the balls as identical, in which case the counting method is different and more complex. Another error is to try to constrain the number of balls per box, even though the question explicitly allows any number. When each distinct object can go independently into any of several distinct boxes, the correct approach is to raise the number of boxes to the power of the number of objects. Final Answer: The total number of distributions is 4096.

Question 14

Two separate sets each contain six different items. In how many ways can we select a collection of items such that at least one item is chosen from each set?

Accepted Answer

Introduction / Context: This question tests your understanding of counting subsets with an “at least one” condition applied to two different groups. We must count all possible selections that include at least one item from the first set and at least one item from the second set. This is a classic application of subtracting unwanted cases from the total number of subsets. Given Data / Assumptions: Set A has 6 distinct items. Set B has 6 distinct items. We select any subset of items from the union of A and B. At least one item must come from A and at least one item from B. Concept / Approach: Without any restrictions, each of the 12 items (6 from A, 6 from B) can either be chosen or not chosen. That gives 2^12 possible subsets. We must subtract the subsets that violate the requirement: subsets with no item from A, subsets with no item from B, and then add back the subset with no items from both sets, since it was subtracted twice. Step-by-Step Solution: Step 1: Total subsets with no restriction = 2^12 = 4096.Step 2: Subsets with no item from A mean we choose only from B. There are 2^6 = 64 such subsets.Step 3: Subsets with no item from B mean we choose only from A. There are also 2^6 = 64 such subsets.Step 4: The empty set (choosing nothing) has been subtracted twice, once in each of the previous two steps, so we must add it back once.Step 5: Required count = 4096 - 64 - 64 + 1 = 4096 - 128 + 1 = 3969. Verification / Alternative check: We can think of independently choosing a nonempty subset from A and a nonempty subset from B and then combining them. The number of nonempty subsets of a 6-element set is 2^6 - 1 = 63. Thus, the total number of valid selections is 63 * 63 = 3969, which confirms our earlier result. Why Other Options Are Wrong: 2856 does not correspond to any simple combination of powers of 2 here and likely reflects partial subtraction of invalid cases. 480 is far too small and might come from mistakenly using combinations like 6C1 or 6C2 instead of considering all subset sizes. 4096 is the total number of subsets with no restriction and fails to enforce the “at least one from each set” condition. Common Pitfalls: Many students forget to add back the empty set when applying the inclusion exclusion principle. Others incorrectly remove subsets that have no items from A or B without thinking about overlaps. A safer approach is often to think in terms of selecting nonempty subsets from each set and multiplying the counts, which correctly handles the condition that both sets contribute at least one element. Final Answer: The number of valid selections is 3969.

Question 15

Sixteen persons including a host are to take their seats in a row for a group photograph. If two particular persons must always sit on either side of the host, in how many different seating arrangements is this possible?

Accepted Answer

Introduction / Context: This is a seating arrangement problem involving a linear row and a special condition about relative positions. We have a host and two particular guests who must always sit immediately on either side of the host. We must count how many different linear arrangements satisfy this constraint when all 16 persons are lined up for a photograph. Given Data / Assumptions: There are 16 persons in total, including the host and two particular guests. Everyone is seated in a single row of 16 seats. The two particular persons must sit immediately to the left and right of the host, in any order. The remaining 13 persons can be seated anywhere in the other seats. Concept / Approach: We treat the host and the two special persons as a block of three adjacent seats, where the host is in the middle and the two special persons occupy the ends. This block can appear in several positions along the row, and the order of the two special persons around the host can also vary. Once we place this block, we arrange the remaining people in the remaining seats. Step-by-Step Solution: Step 1: Consider the three-person block consisting of the host in the middle and the two special guests on either side.Step 2: Within this block, the two special guests can be arranged on the left and right of the host in 2 ways.Step 3: Along the row of 16 seats, this three-seat block can start at positions 1 through 14, so there are 14 possible locations for the block.Step 4: The remaining 13 seats are to be filled by the remaining 13 people. They can be arranged in 13! ways.Step 5: Total arrangements = 14 (positions for the block) * 2 (orders of the two special guests within the block) * 13! (arrangements of the others) = 14! * 2. Verification / Alternative check: Another way is to first place the host (16 choices), then place the two special guests on adjacent sides if possible. However, edge positions complicate counting and lead back to a similar block-based reasoning. The block approach is cleaner and confirms the answer 14! × 2. Why Other Options Are Wrong: 16! × 2 counts all permutations of 16 people and doubles them, ignoring the adjacency constraint around the host. 18! × 2 and 14! do not reflect the correct combination of choices for block placement and internal arrangement of people. Common Pitfalls: Students sometimes forget to consider that the three-person block can slide along the row, or they forget the factor of 2 for the interchange of the two special guests. Others mistakenly treat the situation as circular seating or ignore that the host must be in the middle of the block. Carefully identifying the block and then counting its placements and internal permutations avoids these issues. Final Answer: The number of valid seating arrangements is 14! × 2.

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