Question 1

In the word RAINBOW, how many distinct permutations of all the letters are possible in which R and W occupy the two end positions (in either order)?

Accepted Answer

Introduction / Context: This arrangement problem involves placing specific letters at fixed types of positions, namely the ends of a word. The word RAINBOW has seven distinct letters, and we must count how many permutations have R and W at the two extreme positions, in any order. Once the end letters are fixed, the remaining letters can be permuted freely in the middle positions. Given Data / Assumptions: The word is RAINBOW, with letters R, A, I, N, B, O, W. All letters are distinct. The two end positions are the first and last positions of the seven letter arrangement. R and W must occupy the end positions in any order. The remaining five letters (A, I, N, B, O) fill the middle positions. Concept / Approach: First, determine how many ways R and W can be placed at the two ends. Then, determine how many ways to arrange the remaining five distinct letters in the five middle positions. Multiply these counts to get the total number of valid arrangements. This is a straightforward permutation problem with an additional constraint on specific positions. Step-by-Step Solution: Step 1: Place R and W at the two ends. The two end positions are left and right ends of the word. R can be at the left and W at the right, or W at the left and R at the right. Number of ways to assign R and W to the ends = 2. Step 2: Arrange the remaining letters in the middle positions. Remaining letters: A, I, N, B, O (5 distinct letters). Number of ways to arrange 5 distinct letters = 5! = 120. Step 3: Multiply the contributions. Total valid permutations = 2 * 5! = 2 * 120 = 240. Verification / Alternative check: Instead of thinking in two steps, we can imagine first choosing which letter goes to the left end (2 choices: R or W), which automatically fixes the right end as the other letter. Then we arrange the remaining 5 letters in the middle. This again leads to 2 * 120 = 240 permutations, confirming the result. Why Other Options Are Wrong: Option 120 corresponds to fixing R and W at specific ends but forgetting that they can be swapped. Option 180 or 210 do not match any simple product of 2 and a factorial. Option 360 would be 3 * 120, suggesting that three end configurations were counted, which is not possible here because only R and W can occupy the ends under the given condition. Common Pitfalls: A frequent mistake is to treat the two ends as indistinguishable, counting only 5! = 120 arrangements and ignoring swapping R and W. Another error is to forget that the remaining letters are all distinct and therefore should be arranged using 5!. Explicitly separating the assignment of end letters from the permutations of middle letters avoids such mistakes. Final Answer: There are 240 different permutations of RAINBOW where R and W occupy the two end positions.

Question 2

A college has 10 basketball players available. A team of 5 players is to be selected, and then one of those 5 players is to be chosen as the captain. In how many different ways can the 5 member team and its captain be selected?

Accepted Answer

Introduction / Context: This question is about forming a team and then assigning a special role within that team. We have a pool of 10 distinct players. We first choose a team of 5 players and then select one of those 5 as captain. The team is unordered, but the choice of captain introduces an extra factor because different captains from the same group of players produce different valid outcomes. Given Data / Assumptions: Total available basketball players = 10. Team size required = 5 players. Exactly one captain must be chosen from among the 5 team members. All players are distinct and one person can hold only one role. Concept / Approach: This is a two stage process. First, we select the 5 players who will be on the team. This is a combinations problem because the order of players within the team does not matter. Second, from the chosen 5, we select one player to be captain, which is another simple choice. The total number of outcomes is the product of the number of ways to form the team and the number of ways to assign the captaincy within that team. Step-by-Step Solution: Step 1: Select 5 players out of 10 to form the team. Number of ways = C(10, 5). C(10, 5) = 252. Step 2: From these 5 team members, choose 1 as captain. Number of ways to choose a captain = 5. Step 3: Multiply to get the total number of team plus captain selections. Total number of outcomes = C(10, 5) * 5 = 252 * 5 = 1260. Verification / Alternative check: Another way to see this is to think first of choosing the captain from the 10 players, and then selecting the remaining 4 team members from the remaining 9 players. The number of ways then is 10 * C(9, 4). C(9, 4) = 126, so we get 10 * 126 = 1260, which matches the earlier result. Both approaches yield the same answer, confirming correctness. Why Other Options Are Wrong: The value 210 equals C(10, 4) or C(10, 6) and misses the additional factor for choosing the captain. The numbers 720 and 1512 do not match the combination expression for selecting the team and captain. In particular, 720 is 6!, which is unrelated to this setup, and 1512 is larger than the correct count and might result from overcounting some selections or using permutations incorrectly. Common Pitfalls: A common error is to use permutations directly on 10 players for 5 positions, which would treat every ordering of the team as different even though the team itself is unordered. Another mistake is to forget that the captain must be one of the 5 selected players and either choose the captain from all 10 and then again from the same set or misuse factorials. Clearly separating the selection of the team from the selection of the captain helps avoid these issues. Final Answer: The 5 member team with its captain can be selected in 1260 different ways.

Question 3

In how many different permutations can the letters of the word CORPORATION be arranged so that all the vowels always come together as a single block?

Accepted Answer

Introduction / Context: This arrangement problem asks us to treat all vowels in the word CORPORATION as a single block so that they always stay together. The word contains repeated letters, which means we must use permutations of multiset letters rather than simple factorials. Once we group the vowels into one block, we arrange that block and the consonants, taking into account the repeated letters, and then arrange the vowels inside the block. Given Data / Assumptions: The word is CORPORATION. Letters and frequencies: C(1), O(3), R(2), P(1), A(1), T(1), I(1), N(1). Total letters = 11. Vowels: O, O, O, A, I (5 letters, O repeated 3 times). Consonants: C, R, R, P, T, N (6 letters, R repeated 2 times). All vowels must appear together in every arrangement. Concept / Approach: To keep all vowels together, we think of the vowels as forming a single super letter or block. Then we count permutations of this block together with the consonants. Because there are repeated consonants (two R letters), we divide by the factorial of the repetition count. After finding the number of ways to arrange the block plus consonants, we multiply by the number of ways to arrange the vowels within the block, again dividing by the factorial of the repeated vowel O. The total number of valid permutations is the product of these two factors. Step-by-Step Solution: Step 1: Consider all vowels as a single block V. We have block V plus consonants C, R, R, P, T, N. Total objects to arrange now = 1 (block V) + 6 consonants = 7 objects. Among these, R appears twice. Step 2: Count ways to arrange the 7 objects with repeated R. Number of permutations of these 7 objects = 7! / 2!. 7! = 5040, and 2! = 2, so 7! / 2! = 2520. Step 3: Count ways to arrange vowels inside block V. Vowels: O, O, O, A, I (5 letters with O repeated 3 times). Number of permutations = 5! / 3!. 5! = 120 and 3! = 6, so 5! / 3! = 120 / 6 = 20. Step 4: Multiply these results. Total valid permutations = (7! / 2!) * (5! / 3!) = 2520 * 20 = 50400. Verification / Alternative check: We can check the logic by considering that if there were no repetition, we would simply take 7! for the block and consonants and 5! for the vowels. The repetitions of R and O require division by 2! and 3! respectively, which is exactly what we have done. No other letters are repeated, so no further divisions are needed. The final product 50400 is consistent with a multiset permutation approach. Why Other Options Are Wrong: The value 50000 is close but does not follow from any factorial expression with the correct repetitions. The value 40500 is also not equal to 2520 * 20. The value 5040 corresponds to 7! with no adjustments and ignores vowel arrangements and repetitions. The value 72000 is larger than the correct total and suggests that repetition handling or block arrangement has been misapplied. Common Pitfalls: Students often forget to treat the vowels as a single block initially or ignore some of the repeated letters. Another error is to multiply by 5! instead of 5! / 3!, thereby overcounting arrangements of vowels due to the identical O letters. Similarly, some learners forget to divide by 2! for the repeated R. Always list letter frequencies, treat the vowel block as a single object first, and apply multiset permutation formulas carefully. Final Answer: The letters of the word CORPORATION can be arranged with all vowels together in 50400 different ways.

Question 4

From a total of 7 men and 3 women, in how many different ways can a group of 5 men and 2 women be formed?

Accepted Answer

Introduction / Context: This is a straightforward combinations question involving separate selections from two groups. We must form a group with a fixed number of men and a fixed number of women from given pools of each. Because only the composition of the group matters and not the order, combinations are the appropriate tool for counting such groups. Given Data / Assumptions: Total men available = 7. Total women available = 3. Required group composition: 5 men and 2 women. The group is unordered, so the order of members in the group does not matter. Concept / Approach: We select 5 men from the 7 using combinations and 2 women from the 3 using combinations. Since these choices are independent and can occur together, the total number of groups is the product of the number of ways to choose the men and the number of ways to choose the women. This is a direct application of the multiplication principle for independent selection stages. Step-by-Step Solution: Step 1: Choose 5 men from 7 men. Number of ways = C(7, 5). C(7, 5) = C(7, 2) = 21. Step 2: Choose 2 women from 3 women. Number of ways = C(3, 2) = 3. Step 3: Multiply to get the total number of groups. Total groups = C(7, 5) * C(3, 2) = 21 * 3 = 63. Verification / Alternative check: We can verify by noting that for each of the 21 possible selections of 5 men, there are exactly 3 possible pairs of women. So each men selection leads to 3 distinct full groups, and these products do not overlap because the chosen sets of men differ. Thus 21 sets of men each combine with 3 sets of women, giving 21 * 3 = 63 groups in total. Why Other Options Are Wrong: Values such as 54 or 64 do not equal the product 21 * 3. The option 36 might come from incorrectly choosing 6 people instead of 7 or misusing the combination formula. The number 70 equals C(8, 2) or C(8, 6) and is unrelated to the specific selection of 5 from 7 and 2 from 3. Only 63 matches the correct combination calculations. Common Pitfalls: A common mistake is to try to treat the problem as a single combination from all 10 people, ignoring the requirement that the group must have exactly 5 men and 2 women. Another error is to miscalculate C(7, 5) or C(3, 2). Keeping the male and female selections separate and then multiplying the combination counts helps avoid these errors. Final Answer: The number of ways to form a group of 5 men and 2 women is 63.

Question 5

Using the digits 1, 4, 7, 8 and 9, how many distinct 3 digit numbers can be formed if no digit is repeated in a number?

Accepted Answer

Introduction / Context: This question is about forming 3 digit numbers from a set of distinct digits without repetition. Because all the given digits are non zero, any of them may be used in the hundreds place without worrying about leading zeros. The problem is therefore a simple permutation count of selecting and arranging 3 digits out of 5 distinct digits. Given Data / Assumptions: Available digits: 1, 4, 7, 8, 9. We want to form 3 digit numbers. No digit may repeat within a number. All numbers must be valid 3 digit integers (hundreds digit cannot be zero, but there is no zero here anyway). Concept / Approach: Because every digit is distinct and all can appear in any position, we simply count permutations of 5 digits taken 3 at a time. This equals P(5, 3) and corresponds to choosing a digit for the hundreds place, then a different digit for the tens place, and then a third different digit for the units place. Step-by-Step Solution: Step 1: Consider the hundreds place. There are 5 choices for the first digit (1, 4, 7, 8 or 9). Step 2: Consider the tens place. After choosing the hundreds digit, 4 digits remain available. So there are 4 choices for the tens digit. Step 3: Consider the units place. Now 3 digits remain unused, so there are 3 choices for the units digit. Step 4: Multiply the choices. Total 3 digit numbers = 5 * 4 * 3 = 60. Verification / Alternative check: Using permutation notation, the number of such numbers is P(5, 3) = 5! / (5 - 3)! = 5! / 2!. 5! is 120 and 2! is 2, so P(5, 3) = 120 / 2 = 60. This matches the direct step by step counting method, confirming the result. Why Other Options Are Wrong: The numbers 26, 50 and 64 do not equal the correct permutation value of 60. They may result from confusion with combinations (which would be C(5, 3) = 10) or from partial counts. The value 40 could come from mistakenly restricting the hundreds digit or failing to consider all remaining choices correctly. None of these values corresponds to 5 * 4 * 3. Common Pitfalls: Some learners mistakenly use combinations and compute only the number of ways to choose sets of 3 digits, ignoring order. Others misinterpret the hundreds place restriction when there is no zero among the digits. Remember that for forming numbers where order matters and digits are distinct, permutations are the correct tool. Here, multiplying 5 * 4 * 3 covers all valid arrangements. Final Answer: There are 60 distinct 3 digit numbers that can be formed under the given conditions.

Question 6

From a group of 4 people, in how many different ways can ordered arrangements of 3 people be formed (that is, in how many ways can we select groups of 3 at a time where order matters)?

Accepted Answer

Introduction / Context: This question is directly about permutations. We have 4 distinct people and want to know in how many ways we can form ordered groups of 3 people. Because the order within each group matters, different sequences of the same 3 people are treated as different arrangements. This is a classic example of permutations of n distinct items taken r at a time. Given Data / Assumptions: Total people available = 4. We form groups of size 3. Within each group, order matters (for example, first, second and third positions are distinct). No person is repeated in the same arrangement. Concept / Approach: The number of ordered arrangements of r items from n distinct items is given by the permutation formula P(n, r) = n! / (n - r)!. Here, n = 4 and r = 3. We can apply this formula directly or count the choices position by position, multiplying the number of choices for each position. Step-by-Step Solution: Step 1: Use the permutation formula. P(4, 3) = 4! / (4 - 3)! = 4! / 1!. 4! = 4 * 3 * 2 * 1 = 24. So, P(4, 3) = 24. Step 2: Alternatively, count position by position. For the first position, there are 4 choices. For the second position, after one person is chosen, 3 people remain. For the third position, after two people are chosen, 2 people remain. Multiplying, number of ways = 4 * 3 * 2 = 24. Verification / Alternative check: Listing all permutations explicitly would confirm that there are 24 distinct ordered triples. However, this is not necessary once we realise that the permutation formula and the direct positional counting both agree on 24. Both methods are logically equivalent and consistent with basic permutation principles. Why Other Options Are Wrong: The value 16 would correspond to 4^2, which is not relevant here. The number 20 might be mistakenly obtained by using combinations (C(4, 3) = 4) and then multiplying incorrectly. The value 36 is larger than 4 * 3 * 2 and would require an extra factor that does not exist in this setting. The value 12 is exactly half of the correct answer and could result from stopping the multiplication early. Common Pitfalls: A common mistake is to use combinations, which count unordered selections, rather than permutations. Another error is to think that there are only C(4, 3) = 4 groups and forget that each group can be arranged in 3! = 6 ways, giving 4 * 6 = 24 ordered arrangements. Keeping the distinction between order sensitive and order insensitive problems clear helps avoid such confusion. Final Answer: There are 24 different ordered arrangements of 3 people from the group of 4.

Question 7

From a set of 6 distinct items, in how many different ways can ordered arrangements of 4 items be formed (that is, in how many ways can we select groups of 4 at a time where order matters)?

Accepted Answer

Introduction / Context: This problem is another example of permutations, where we must form ordered arrangements from a larger set. We have 6 distinct items and wish to create sequences of length 4 using these items without repetition. Because order matters, each different ordering of the same 4 items counts as a different arrangement. Given Data / Assumptions: Total items available = 6. We form ordered groups of size 4. No item is repeated within a single arrangement. Different orders of the same 4 items are counted as different arrangements. Concept / Approach: The number of ordered arrangements of r items chosen from n distinct items is given by the permutation formula P(n, r) = n! / (n - r)!. Here, n = 6 and r = 4. We can plug these directly into the formula or think in terms of choices for each position in the ordered arrangement and multiply the number of choices. Step-by-Step Solution: Step 1: Apply the permutation formula. P(6, 4) = 6! / (6 - 4)! = 6! / 2!. 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 2! = 2. So P(6, 4) = 720 / 2 = 360. Step 2: Alternatively, count by positions. First position: 6 choices. Second position: 5 choices (after one item is used). Third position: 4 choices. Fourth position: 3 choices. Total arrangements = 6 * 5 * 4 * 3 = 360. Verification / Alternative check: The two methods 6! / 2! and 6 * 5 * 4 * 3 agree, both giving 360. Listing all permutations is possible but unnecessary and time consuming. The standard permutation formula is sufficient to verify that 360 is the correct count for the number of ordered arrangements of 4 items from 6 distinct items. Why Other Options Are Wrong: The value 720 equals 6! and would correspond to arranging all 6 items, not just 4 at a time. The value 640 does not result from any natural factorial based calculation here. The value 740 is close to 720 but is not a factorial or permutation count. The number 120 equals 5! and is too small for P(6, 4). None of these values matches 6 * 5 * 4 * 3. Common Pitfalls: A common mistake is to treat the problem as combinations, counting only C(6, 4) groups and ignoring order. Another error is to miscalculate 6! or forget to divide by 2!. Some might also incorrectly think that all 6 items must be used and use 6! directly. Being clear that we want ordered groups of length 4 and applying P(6, 4) = 6! / 2! ensures the correct approach. Final Answer: There are 360 different ordered arrangements of 4 items from the set of 6.

Question 8

From a set of 20 distinct objects, how many different groups of 5 objects can be formed if the order of objects inside each group does not matter at all?

Accepted Answer

Introduction / Context: This question tests your understanding of basic combinatorics, specifically the concept of combinations. When we form groups from a larger set and do not care about the internal order of items in each group, we use combinations, not permutations. Such questions are very common in aptitude exams, engineering entrance tests, and competitive exams, so it is important to clearly know when to use nCr instead of nPr. Given Data / Assumptions: There are 20 distinct objects in total. We must choose groups containing exactly 5 objects. The order of objects within each group does not matter. Each object can either be in a group or not, and no object is repeated inside a group. Concept / Approach: To count selections where order does not matter, we use combinations. The formula for combinations is: nCr = n! / (r! * (n - r)!) Here n = 20 and r = 5. We simply substitute in this formula and simplify. This gives the exact number of different 5-object groups that can be formed from 20 distinct objects. Step-by-Step Solution: Step 1: Identify n = 20 (total objects) and r = 5 (objects in one group). Step 2: Use the formula nCr = n! / (r! * (n - r)!). Step 3: Compute 20C5 = 20! / (5! * 15!). Step 4: Expand only what is needed: 20C5 = (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1). Step 5: Simplify the numerator and denominator carefully: (20 * 19 * 18 * 17 * 16) = 1860480 and 5! = 120. Step 6: Divide: 1860480 / 120 = 15504. Step 7: Therefore, the number of distinct 5-object groups is 15504. Verification / Alternative check: We can quickly check reasonableness. 20C1 is 20, 20C2 is 190, and combinations grow larger as r increases up to around the middle. Since 20C5 is symmetric with 20C15 and is near the central region, a value in the tens of thousands is reasonable. Also, a calculator or combinatorics table confirms that 20C5 = 15504, so our result is consistent with standard values. Why Other Options Are Wrong: 57090: This does not match the correct application of the combination formula and is too large. 15540: Very close in magnitude but results from incorrect simplification or arithmetic. 23670: Again, this does not correspond to 20C5 and comes from an incorrect calculation. These distractors are typical of small arithmetic mistakes in factorial simplification. Common Pitfalls: Many students confuse permutations and combinations. Using nPr instead of nCr would wrongly treat different orders of the same group as different arrangements. Another frequent mistake is partial cancellation of factorial terms, leading to minor numerical errors like 15540 instead of 15504. It is important to carefully cancel common terms in numerator and denominator and to compute step by step for accuracy. Also, forgetting that order does not matter here leads to overcounting and a much higher incorrect number. Final Answer: The total number of distinct groups of 5 objects that can be formed from 20 different objects, when order inside each group does not matter, is 15504.

Question 9

From a committee of 17 people, in how many different ways can you select a subcommittee of 7 members if only the selection matters and the order of members does not matter?

Accepted Answer

Introduction / Context: This problem is a direct application of combinations in combinatorics. We are forming a smaller subcommittee from a larger committee, and only the set of people in the subcommittee matters. Their seating order or any ranking is not relevant at this stage. Such selection problems are extremely common in aptitude tests, management entrance exams, and interview questions that test logical thinking and basic mathematics. Given Data / Assumptions: Total number of people on the main committee: 17. Number of people required in the subcommittee: 7. All 17 people are distinct individuals. The order in which the 7 people are chosen does not matter. No person can be chosen more than once. Concept / Approach: When we select r people from n distinct people and we only care about which people are chosen (not the order), we use combinations. The formula is: nCr = n! / (r! * (n - r)!) Here, n = 17 and r = 7. We simply compute 17C7 using this formula. This gives the number of unique 7-member groups that can be formed from the 17 members. Step-by-Step Solution: Step 1: Identify n = 17 and r = 7. Step 2: Use the combination formula 17C7 = 17! / (7! * 10!). Step 3: Write 17! / 10! as 17 * 16 * 15 * 14 * 13 * 12 * 11. Step 4: Compute numerator: 17 * 16 * 15 * 14 * 13 * 12 * 11. Step 5: Divide this product by 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1. Step 6: Carefully cancel common factors and simplify step by step to avoid mistakes. Step 7: After simplification, we get 17C7 = 19448. Verification / Alternative check: We can verify the answer by using the symmetry property of combinations. Note that 17C7 = 17C10 because choosing 7 members to include is equivalent to choosing 10 members to exclude. Many scientific calculators can compute 17C7 directly and confirm the result 19448. Checking with any reliable combinatorics table or tool also matches this value, which increases confidence in the result. Why Other Options Are Wrong: 19000: A rounded-looking number but not the result of 17C7. 19821: Numerically close but does not match the exact combination formula result. 19340: Another nearby value that likely arises from an arithmetic error in cancellation. Only 19448 satisfies the exact combination computation for 17C7. Common Pitfalls: Students often confuse combinations with permutations and may wrongly compute 17P7. That would be much larger and completely wrong here. Another common error is incorrect simplification of factorials due to skipping steps or doing mental arithmetic too quickly. Writing out cancellation clearly and using the combination formula carefully helps avoid these mistakes. It is also important not to treat 17C7 as 17^7, which is a completely different operation. Final Answer: The number of different 7-member subcommittees that can be selected from a committee of 17 people is 19448.

Question 10

From a standard deck of 52 distinct playing cards, how many different 5-card hands can be drawn if the order in which the cards are dealt does not matter?

Accepted Answer

Introduction / Context: This is a classic card combinatorics problem that appears very frequently in probability and aptitude topics. When dealing with poker hands or any 5-card hands, we consider only which cards are in the hand, not the order in which they were drawn. Understanding how to count such hands using combinations is fundamental for more advanced probability questions involving cards, suits, and ranks. Given Data / Assumptions: A standard deck has 52 distinct cards. We are drawing a hand of 5 cards. The order of cards in the hand does not matter. No card is repeated since a standard deck does not allow duplicates. Every 5-card subset is treated as one unique hand. Concept / Approach: Because order does not matter, we again use combinations. We must compute the number of ways to choose 5 cards out of 52. This is written as 52C5. The formula is: nCr = n! / (r! * (n - r)!) So for this problem, 52C5 = 52! / (5! * 47!). Step-by-Step Solution: Step 1: Identify n = 52 and r = 5. Step 2: Apply the combination formula: 52C5 = 52! / (5! * 47!). Step 3: Write 52! / 47! as 52 * 51 * 50 * 49 * 48. Step 4: Compute numerator product: 52 * 51 * 50 * 49 * 48. Step 5: Divide by 5! = 120, simplifying step by step to reduce arithmetic mistakes. Step 6: After careful simplification, the result is 2598960. Step 7: Therefore, there are 2598960 distinct 5-card hands. Verification / Alternative check: This value 2598960 is a standard result in probability theory and card combinatorics. It is widely quoted in textbooks and online resources for the number of 5-card hands from a deck of 52 cards. You can also verify the value using a scientific calculator with nCr functionality. Any deviation from 2598960 indicates an arithmetic or formula error. Why Other Options Are Wrong: 2589860: Close in magnitude but not equal to 52C5. 2598970: Only slightly different in the last digits, indicating a small calculation error. 2430803: Significantly smaller and not linked to the correct combination computation. Only 2598960 exactly matches the mathematically correct value for 52C5. Common Pitfalls: Common mistakes include treating each arrangement as different by using permutations instead of combinations, which would dramatically overcount the number of hands. Another mistake is performing the factorial division incorrectly, especially when handling large numbers. Breaking the calculation into manageable steps and cancelling terms before multiplying avoids overflow and reduces the chance of arithmetic error. Remember that for card hands, unless explicitly stated otherwise, order never matters, so combinations are the correct tool. Final Answer: The total number of distinct 5-card hands that can be drawn from a 52-card deck is 2598960.

Question 11

A school has scheduled 3 volleyball games, 2 soccer games, and 4 basketball games. If you have a ticket that allows you to attend exactly 3 games and you decide to watch 2 basketball games and 1 game from the other sports, in how many different ways…

Accepted Answer

Introduction / Context: This question mixes simple combinatorics with a real life scenario about choosing sports events. It tests your ability to correctly interpret phrases like "2 basketball games and 1 of the other events" and then apply combinations to count choices from different categories. Understanding how to handle selections from separate groups is important for many exam problems in which items are classified by type. Given Data / Assumptions: Number of volleyball games: 3. Number of soccer games: 2. Number of basketball games: 4. You must attend exactly 3 games in total. Your choice must include 2 basketball games and exactly 1 non-basketball game (volleyball or soccer). Each game is distinct. Concept / Approach: We select games from two separate groups: First, choose 2 basketball games out of 4 available. Second, choose 1 game from the 3 volleyball and 2 soccer games, that is, from 5 non-basketball games. In both cases, the order in which you choose games does not matter. Therefore, combinations are used for each selection, and then the counts are multiplied using the fundamental principle of counting. Step-by-Step Solution: Step 1: Number of ways to choose 2 basketball games from 4 is 4C2. Step 2: Compute 4C2 = 4! / (2! * 2!) = 6. Step 3: Total non-basketball games = 3 volleyball + 2 soccer = 5. Step 4: Number of ways to choose 1 non-basketball game from 5 is 5C1 = 5. Step 5: Use multiplication principle: total ways = 4C2 * 5C1 = 6 * 5. Step 6: Compute 6 * 5 = 30. Step 7: Therefore, there are 30 different ways to attend exactly 2 basketball games and 1 other game. Verification / Alternative check: You can list combinations more systematically in a smaller toy example to verify the structure and then trust the formula. For instance, if there were 3 basketball games and 2 others, you could manually list selections to see that the pattern 3C2 * 2C1 works correctly. Scaling back to the real numbers (4 and 5) keeps the logic identical, so the multiplication of separate combination counts is justified. Why Other Options Are Wrong: 25: This might come from incorrectly doing 5C2, ignoring the condition about exactly 2 basketball games. 50: Could arise from mistakenly adding or multiplying extra factors without proper reasoning. 75: Too large and does not correspond to a valid combination structure for the given conditions. Only 30 matches the correct breakdown 4C2 * 5C1. Common Pitfalls: A common mistake is to combine all games and choose any 3 without enforcing the condition of exactly 2 basketball games. Another error is to treat volleyball and soccer separately and double count mixed selections. Always separate the problem into independent choices (here, basketball choice and non-basketball choice) and then use combinations for each selection. Finally, students sometimes confuse 5C1 with 5C2 or misapply permutations where the sequence of attending games is treated as important, which it is not in this question. Final Answer: The number of different ways to watch exactly two basketball games and one other game is 30.

Question 12

In a standard deck of cards, how many distinct 5-card hands form a straight using the ranks 4, 5, 6, 7, and 8 (that is, starting with 4 and ending with 8), if suits can be chosen freely and the order of cards in the hand does not matter?

Accepted Answer

Introduction / Context: This question is about counting specific poker type hands: 5-card straights that run from 4 to 8. The focus is on how many distinct hands can be made with those exact ranks while suits can vary freely. This is a typical intermediate card combinatorics question where ranks are fixed but suits are flexible. Understanding such counting is helpful for probability problems involving straights and related events in card games. Given Data / Assumptions: We use a standard 52-card deck with 4 suits. The straight must use the ranks 4, 5, 6, 7, and 8. For each rank, any of the four suits (clubs, diamonds, hearts, spades) is allowed. The order of cards in the hand does not matter; hands are just sets of cards. No card can repeat, so each chosen card has a distinct combination of rank and suit. Concept / Approach: The ranks are fixed: we must take exactly one card of rank 4, one of rank 5, one of rank 6, one of rank 7, and one of rank 8. For each rank, there are 4 possible suits. The choices for different ranks are independent of each other. Therefore, the total number of distinct hands is the product of suit choices across all five ranks. Step-by-Step Solution: Step 1: For rank 4, there are 4 possible cards (one in each suit). Step 2: For rank 5, there are again 4 possible cards. Step 3: Similarly, ranks 6, 7, and 8 each contribute 4 possible cards. Step 4: Because we choose exactly one card of each rank, multiply the choices: 4 * 4 * 4 * 4 * 4. Step 5: Compute 4^5 = 4 * 4 * 4 * 4 * 4 = 1024. Step 6: The order of cards does not matter, but each 5-card collection here is already a unique hand because the composition (specific rank and suit of each card) is different. Step 7: Thus, total number of such straight hands is 1024. Verification / Alternative check: We can think of this as a function assignment: for each of the 5 distinct ranks, choose one of 4 suits, which gives 4^5 possibilities. There is no overlap, since each resulting set of cards is unique. No division by 5! is required because we are not counting sequences; we are counting distinct sets of cards already specified by their ranks. This reasoning confirms 1024 as the correct answer. Why Other Options Are Wrong: 1296: Equal to 6^4, which does not relate correctly to the structure of this problem. 1094: A random looking number, not a power or simple product relevant to the scenario. 1200: Does not arise from any valid combination or power count for five ranks with four suits. None of these are equal to 4^5, which is the correct computation. Common Pitfalls: A typical mistake is to treat this as 52C5, which would count all possible 5-card hands instead of only straights from 4 to 8. Another error is to think that one must divide by 5! to account for order, forgetting that by directly counting suit assignments per rank we already avoid overcounting again. Some learners also confuse this with counting the number of different rank sequences, but here the rank pattern is completely fixed, so we only vary suits. Final Answer: The number of distinct 5-card hands that form a straight using ranks 4, 5, 6, 7, and 8 is 1024.

Question 13

In a marathon race with 50 runners, you want to predict exactly which runner will finish first, which runner will finish second, and which runner will finish third. In how many different ways can the top three positions occur?

Accepted Answer

Introduction / Context: This problem is about ordered arrangements, which are handled using permutations. We are not just choosing which three runners are in the top three, but also specifying the exact finishing order: first, second, and third. This makes the problem more than a simple combination. Understanding when to switch from combinations to permutations is crucial in many real life ranking and arrangement problems. Given Data / Assumptions: Total number of runners in the marathon: 50. We are concerned with the first three finishing positions: first, second, and third. All 50 runners are distinct individuals. The order of finish matters, since first place is different from second or third. No ties are considered; exactly one runner holds each of the top three positions. Concept / Approach: Because we care about both who is in the top three and the exact order, we use permutations rather than combinations. The number of ways to choose and arrange r items from n distinct items is given by: nPr = n * (n - 1) * (n - 2) * ... up to r factors. Here, n = 50 and r = 3. So we must compute 50P3, which counts all ordered triples of distinct runners. Step-by-Step Solution: Step 1: There are 50 choices for who comes first. Step 2: After choosing the first place runner, 49 runners remain, so there are 49 choices for second place. Step 3: After that, 48 runners remain, giving 48 choices for third place. Step 4: Apply the multiplication principle: total ways = 50 * 49 * 48. Step 5: Compute 50 * 49 = 2450. Step 6: Then multiply 2450 * 48 = 117600. Step 7: Therefore, there are 117600 possible ordered outcomes for the top three positions. Verification / Alternative check: We can express the same reasoning using the permutation notation 50P3. The formula for permutations is 50P3 = 50! / (50 - 3)!. That gives 50! / 47!, which simplifies to 50 * 49 * 48, the same as our stepwise counting. A calculator that supports nPr will also confirm that 50P3 = 117600, validating the result. Why Other Options Are Wrong: 19600: This is equal to 50C3 and would be correct only if order did not matter. 125000: Looks like 50^3, which would allow repetition, not valid in a race with unique finishers. 60000: A random large number without a correct combinatorial interpretation for this scenario. Only 117600 correctly counts ordered distinct top three outcomes from 50 runners. Common Pitfalls: A very common mistake is to use combinations, 50C3, instead of permutations. That counts only which three runners appear in the top three, ignoring their finish order. Another error is to miscalculate 50 * 49 * 48 due to arithmetic slips. Breaking the multiplication into two steps and checking results with a calculator helps avoid such mistakes. Also, including repetitions such as allowing the same runner to appear multiple times in the top three is logically wrong and inflates the count incorrectly. Final Answer: The number of different ways in which the top three positions can be filled by 50 runners is 117600.

Question 14

When writing all integers from 1 to 1000 inclusive, how many times in total does the digit 7 appear across all these numbers?

Accepted Answer

Introduction / Context: Digit counting questions are popular in aptitude exams because they test both logical reasoning and systematic counting skills. Instead of manually writing and checking all numbers from 1 to 1000, we look for a clean method that uses symmetry and positional analysis. The goal here is to count how many times the digit 7 appears in all positions (units, tens, hundreds) when listing numbers from 1 up to 1000 inclusive. Given Data / Assumptions: Numbers considered: 1, 2, 3, ..., 1000. We count every occurrence of the digit 7 in any position. The number 1000 does not contain the digit 7. Leading zeros are not written, but we may imagine numbers as three digit strings with leading zeros (from 001 to 999) for easier counting. All standard decimal representations are used without skipping any number. Concept / Approach: A common approach is to treat 001 to 999 as three digit numbers and analyze each position (hundreds, tens, units) separately. Due to symmetry, the digit 7 appears equally often in each position. Once we know how many times 7 appears in one position, we can multiply by 3 to get the total from 001 to 999. Finally, we check whether 1000 adds any extra 7, which it does not. Step-by-Step Solution: Step 1: Consider numbers from 000 to 999 as all three digit strings. Step 2: There are 1000 such strings (from 000 to 999). Step 3: For the units place, each digit from 0 to 9 appears equally often. Step 4: Therefore, digit 7 appears 1000 / 10 = 100 times in the units place. Step 5: The same logic applies to the tens place: 7 appears 100 times in the tens position. Step 6: Similarly, in the hundreds place, 7 appears 100 times. Step 7: Total occurrences from 000 to 999 = 100 + 100 + 100 = 300. Step 8: Now ignore 000 (not part of 1 to 1000). This does not affect the count of digit 7. Step 9: Check 1000, which has no 7, so the total remains 300. Verification / Alternative check: You can make a quick partial check by counting 7s from 1 to 99 and seeing the pattern. From 0 to 99, the digit 7 appears 20 times in the units place and 20 times in the tens place for a total of 40. Extending this logic up to 999 results in 300 total occurrences, which matches our positional reasoning. Computer based checks also confirm this total. Why Other Options Are Wrong: 243: This corresponds to 3 * 9^2, which is not related to the correct equal distribution logic. 301: Just one more than the correct value, probably from miscounting 1000 as containing a 7 somewhere. 290: A rounded guess that ignores the exact distribution across all positions. Only 300 fits the precise positional counting argument. Common Pitfalls: A frequent error is to try to count manually in ranges and accidentally double count or skip numbers. Another mistake is forgetting about the hundreds place or misinterpreting leading zeros. Some students mistakenly think that each digit appears only 90 times instead of 100 in one position due to confusion between ranges. Using the structured approach of considering 000 to 999 and then adjusting for 1 to 1000 makes the problem much easier and less error prone. Final Answer: The digit 7 is written a total of 300 times when listing all integers from 1 to 1000 inclusive.

Question 15

A team of 8 students is going on an excursion using two cars. One car has 5 seats and the other car has 4 seats. In how many different ways can the 8 students be divided between the two cars, assuming seats within a car are not distinguished?

Accepted Answer

Introduction / Context: This question involves distributing people into groups with capacity constraints. Instead of focusing on arranging people in specific seats, we only care about which students go in which car. Such distribution problems come up frequently in combinatorics and in real life planning scenarios. Understanding how to split a group into subgroups according to given size limits is a key counting skill. Given Data / Assumptions: Total number of students: 8. Car 1 has 5 seats. Car 2 has 4 seats. There will be exactly 8 students travelling, so one seat will remain empty. Students are distinct individuals. We do not distinguish individual seats inside a car, only which students are in each car. Concept / Approach: We must count all valid ways of dividing 8 distinct students between two cars with capacities 5 and 4. Let the number of students in the 5 seat car be x and in the 4 seat car be y. We have two constraints: x + y = 8 (all students travel). 0 ≤ x ≤ 5 and 0 ≤ y ≤ 4 (capacity limits). Solving these, only two valid distributions exist: (x, y) = (4, 4) and (5, 3). We then count combinations for each and add them. Step-by-Step Solution: Step 1: Determine possible splits: x + y = 8 with x ≤ 5 and y ≤ 4. Step 2: Valid splits are (4, 4) and (5, 3). No other combination fits both constraints. Step 3: For (4, 4): choose 4 students from 8 for the first car. Number of ways = 8C4. Step 4: Compute 8C4 = 70. The remaining 4 automatically go in the second car. Step 5: For (5, 3): choose 5 students from 8 for the 5 seat car. Number of ways = 8C5. Step 6: Compute 8C5 = 56. The remaining 3 go in the 4 seat car. Step 7: Total ways = 70 + 56 = 126. Verification / Alternative check: We can see that 8C4 = 8C4 and 8C5 = 8C3 by symmetry of combinations. So the total 8C4 + 8C3 = 70 + 56 is consistent with the counting of different ways to pick which students use which car under the given capacity constraints. There is no double counting, since each split has a unique number of students in each car. Why Other Options Are Wrong: 240: Might come from multiplying 8C4 by 2 incorrectly or including invalid distributions. 120: Close to 5!, but this does not match the exact counting logic here. 260: A random sum that does not align with any correct combination breakdown. Only 126 correctly matches the sum of valid distributions (4, 4) and (5, 3). Common Pitfalls: Some students ignore car capacities and assume any split is acceptable, which leads to overcounting. Others mistakenly treat the two cars as indistinguishable, which would require dividing by 2 in some symmetric cases, but here the cars have different capacities, so they are inherently distinct. Finally, it is important not to attempt to assign specific seats inside each car, as the problem only asks for distributions by car, not seat level arrangements. Final Answer: The number of different ways in which the 8 students can travel in the two cars is 126.

Question 16

On the board of a club there are 8 first class petty officers and 6 second class petty officers. In how many different ways can a president, vice president, secretary, and treasurer be chosen if the president and secretary must both be first class p…

Accepted Answer

Introduction / Context: This question tests your ability to count arrangements when positions have specific eligibility constraints. We are choosing four different posts, each from a restricted subset of people. Because the order of posts clearly matters (president is not the same as secretary), we must use permutations inside restricted groups. These kinds of problems are common in exam questions that mix combinatorics and logical thinking. Given Data / Assumptions: First class petty officers: 8 people. Second class petty officers: 6 people. Positions: president, vice president, secretary, treasurer. President and secretary must be first class petty officers. Vice president and treasurer must be second class petty officers. No person can hold more than one post simultaneously. Concept / Approach: We break the problem into two independent parts: Assign president and secretary from the 8 first class petty officers. Assign vice president and treasurer from the 6 second class petty officers. Within each group, the positions are distinct, so we use permutations. Finally, we multiply the counts from both groups because choosing first class posts and second class posts are independent tasks. Step-by-Step Solution: Step 1: Number of ways to choose a president from 8 first class officers = 8. Step 2: After choosing the president, 7 first class officers remain for the secretary post. Step 3: Ways to assign president and secretary from first class officers = 8 * 7 = 56. Step 4: Now consider second class officers for vice president and treasurer. Step 5: Number of ways to choose a vice president from 6 second class officers = 6. Step 6: After that, 5 officers remain for treasurer, so ways = 6 * 5 = 30. Step 7: Total ways = ways to assign first class posts * ways to assign second class posts = 56 * 30 = 1680. Verification / Alternative check: The same calculation can be expressed as permutations: for first class posts, 8P2 = 8 * 7 = 56, and for second class posts, 6P2 = 6 * 5 = 30. The total number of assignments is 8P2 * 6P2 = 56 * 30 = 1680. Both approaches match perfectly, confirming the result is consistent and correct. Why Other Options Are Wrong: 1500: Does not match the exact product 56 * 30 and indicates a miscalculation. 1860: Another incorrect multiplication, possibly from mixing 8C2 or 6C2 with permutations. 1640: Close but still incorrect; it does not arise from any valid combination of permutations here. Only 1680 correctly reflects the separate permutations for first class and second class roles. Common Pitfalls: Some learners treat the four posts as identical, incorrectly using combinations instead of permutations. Others ignore the class restrictions and choose all four officers from the combined pool without distinguishing first class and second class categories. It is also easy to forget that once a person is selected for one role, they cannot be chosen again for another, leading to overcounting. Always carefully interpret role constraints and apply permutations where order (which post is held) clearly matters. Final Answer: The number of different ways to choose the four officers under the given conditions is 1680.

Question 17

In a group of 20 persons there are exactly 2 brothers. In how many different circular arrangements of all 20 persons around a round table will there be exactly one person sitting between the two brothers?

Accepted Answer

Introduction / Context: This problem involves circular permutations with a restriction on the relative positions of two specific people. Circular arrangement questions are more subtle than linear arrangements because rotations of a configuration are considered the same. Here, we want exactly one person between the two brothers, so we must respect both circular symmetry and the spacing constraint. Such questions appear in higher level aptitude and reasoning tests. Given Data / Assumptions: Total persons: 20, including 2 brothers and 18 other distinct people. They are seated around a circular table. Arrangements that differ only by rotation are considered the same circular arrangement. There must be exactly one person between the two brothers. The direction (clockwise or anticlockwise) around the table is fixed for counting, but we account for circular symmetry properly. Concept / Approach: For circular permutations of n distinct people, the number of unrestricted arrangements is (n - 1)!. To handle the constraint, we use a fixing technique. We fix the position of one brother to break rotational symmetry, then decide positions for the second brother consistent with exactly one person between them, and finally permute the remaining 18 people freely in the remaining seats. Step-by-Step Solution: Step 1: Fix Brother 1 at a reference position on the circle to remove rotational duplicates. Step 2: With 20 seats in total, once Brother 1 is fixed, there are 19 remaining seats. Step 3: Brother 2 must sit with exactly one person between him and Brother 1. Step 4: On a circle, there are exactly two such seats for Brother 2: one two steps clockwise from Brother 1 and one two steps anticlockwise. Step 5: Therefore, there are 2 choices for Brother 2 once Brother 1 is fixed. Step 6: After seating both brothers, 18 remaining distinct people can be arranged in the remaining 18 seats in 18! ways. Step 7: Total valid arrangements = 2 * 18!. Verification / Alternative check: If there were no restriction, the number of circular arrangements of 20 distinct people would be 19!. Our answer 2 * 18! is equal to (2 / 20) * 19!, so about one tenth of all unrestricted circular arrangements satisfy the condition that the brothers have exactly one person between them. This ratio is plausible since there are many possible relative positions for two people around a large circle, and only two of them produce exactly one person between them, once rotational symmetry is accounted for. Why Other Options Are Wrong: 18! x 19!: This multiplies two factorial counts incorrectly and does not reflect circular symmetry. 2 x 19!: This would be correct if we ignored circular fixing and overcounted rotations. 18! x 18!: A product with no correct combinatorial interpretation for this setup. Only 2 x 18! correctly follows from fixing one brother, choosing two possible seats for the other, and permuting the rest. Common Pitfalls: A major pitfall is forgetting that circular arrangements have rotational symmetry, which should be handled by fixing one person in place. Another is miscounting the number of valid positions for the second brother; some students think there are more than two such seats without visualizing the circle properly. It is also easy to accidentally use 19! directly rather than 2 * 18!, or to treat the arrangement as linear, which changes the count dramatically. Drawing a rough circle with labeled positions can help clarify the reasoning. Final Answer: The total number of circular arrangements in which exactly one person sits between the two brothers is 2 x 18!.

Question 18

The word MISSISIPPI has 10 letters, with repeated letters I, S, and P. In how many distinct permutations can all the letters of the word MISSISIPPI be arranged?

Accepted Answer

Introduction / Context: This problem is about counting permutations of letters when some letters are repeated. When all letters are distinct, we simply use n! arrangements. However, repeated letters cause overcounting because swapping identical letters does not create a new distinct arrangement. Such problems appear frequently with words like MISSISSIPPI, BALLOON, and so on in combinatorics and aptitude tests. Given Data / Assumptions: Word: MISSISIPPI. Total letters: 10. Letter counts: M appears 1 time, I appears 4 times, S appears 3 times, P appears 2 times. All permutations use all 10 letters at once. Two arrangements that look identical when written are considered the same arrangement. Concept / Approach: The formula for distinct permutations of a multiset with repeated elements is: Total permutations = n! / (n1! * n2! * n3! * ...) where: n is the total number of letters. n1, n2, n3, ... are the frequencies of each distinct letter. For MISSISIPPI, n = 10, with counts 4, 3, 2, and 1 for I, S, P, and M respectively. Step-by-Step Solution: Step 1: Compute the total factorial of letters: 10!. Step 2: Identify repeated letter counts: I = 4, S = 3, P = 2, M = 1. Step 3: Write the formula: total distinct permutations = 10! / (4! * 3! * 2! * 1!). Step 4: Compute 10! = 3628800. Step 5: Compute 4! = 24, 3! = 6, 2! = 2, and 1! = 1. Step 6: Multiply the denominator: 24 * 6 * 2 * 1 = 288. Step 7: Divide 3628800 by 288 to get 12600. Verification / Alternative check: We can simplify division stepwise: 3628800 / 24 = 151200, then 151200 / 6 = 25200, then 25200 / 2 = 12600. This confirms that all cancellations are performed correctly. Also, if all letters were distinct, we would have 10! = 3628800 arrangements, so having 12600 distinct permutations after accounting for repetitions is reasonable, because many permutations collapse to the same visual arrangement when identical letters are swapped. Why Other Options Are Wrong: 12400 and 11160: Both are close but smaller than 12600 and arise from incorrect partial division or forgetting a factorial factor. 16200: Larger than the correct answer and suggests incomplete adjustment for repetition. Only 12600 matches the correct multiset permutation formula and accurate computation. Common Pitfalls: A common error is to ignore repeated letters entirely and just compute 10!, which greatly overestimates the count. Another mistake is to divide by only some of the repeated factorials, for example by 4! and 3! but forgetting 2!. Finally, mixing up the exponents or miscounting the number of occurrences of each letter leads to wrong denominators. Carefully counting each letter and applying the full formula avoids these problems. Final Answer: The total number of distinct permutations of the letters in MISSISIPPI is 12600.

Question 19

You have 5 identical letters A (so the multiset of letters is AAAAA). In how many different ways can you choose zero or more letters from this set, where selections that differ only in how many As are chosen are considered distinct?

Accepted Answer

Introduction / Context: This problem focuses on counting selections when all items are identical. Unlike typical combination questions with distinct objects, here each letter is the same (all are A). Therefore, the only thing that matters is how many As are chosen, not which particular ones. Such questions are good stepping stones towards understanding combinations with identical objects and basic ideas of distribution. Given Data / Assumptions: There are 5 identical letters A. We may select zero or more letters from these 5. Selections are distinguished only by the count of As, not by particular positions. We do not care about order; a group of 3 As is the same no matter which physical As are chosen. Concept / Approach: Because the letters are identical, all that matters is the number of letters selected. Possible choices are: Choose 0 As. Choose 1 A. Choose 2 As. Choose 3 As. Choose 4 As. Choose 5 As. Each different count corresponds to a unique selection. Therefore, we simply count the possible counts from 0 up to 5 inclusive. Step-by-Step Solution: Step 1: List all possible numbers of As you can select: 0, 1, 2, 3, 4, 5. Step 2: Each of these corresponds to a distinct selection, because the letters are identical. Step 3: There are 6 possibilities in this list. Step 4: Therefore, the number of different selections of zero or more letters from AAAAA is 6. Verification / Alternative check: This can be interpreted as the number of integer solutions to k between 0 and 5 inclusive, where k is the number of As selected. Since k can take exactly 6 values (0 through 5), the count of distinct selections is 6. There is no other structure such as order or labeling that changes this result, so the count is robust to any rephrasing of the problem. Why Other Options Are Wrong: 5: This would correspond to choosing at least 1 A, but the problem explicitly allows zero letters. 2: This might come from mistakenly thinking there are only two possibilities, selecting or not selecting, which ignores the different counts of As. 8: Larger than the number of possible counts; there are only 6 distinct counts from 0 to 5. Only 6 matches the correct count of allowed selection sizes. Common Pitfalls: Students sometimes treat the As as if they were distinct, leading to unnecessary combinatorial expressions. Others forget that selecting zero letters is a valid option and undercount by one. The key is to recognize that identical objects reduce a selection problem to counting how many items are chosen, rather than which specific items. Carefully reading the phrase "zero or more" ensures that the empty selection is included. Final Answer: The number of different ways to select zero or more letters from AAAAA is 6.

Question 20

In a horse race with 6 distinct horses, two horses are named Goldenrod and No Hope. Assuming all 6 horses finish the race with no ties, in how many different possible finishing orders will No Hope finish somewhere ahead of Goldenrod?

Accepted Answer

Introduction / Context: This is a classic relative order problem involving permutations with a condition. We have 6 horses and are interested in the finishing order, but with the added restriction that one specific horse, No Hope, must finish before another specific horse, Goldenrod. These types of questions help build intuition for symmetry in permutations and relative ordering constraints. Given Data / Assumptions: Total horses: 6, all distinct. Two special horses: No Hope (N) and Goldenrod (G). All horses finish the race with no ties. We consider all possible full finishing orders of the 6 horses. We only count those orders where N finishes somewhere before G. Concept / Approach: Without any restriction, all 6 horses can finish in any order, giving 6! possible permutations. Now, observe symmetry: in the set of all possible permutations, the relative positions of N and G are equally likely to be N before G or G before N. Therefore, exactly half of all permutations will have N before G, and the other half will have G before N. This symmetry argument leads to a very quick solution. Step-by-Step Solution: Step 1: Compute the total number of unrestricted finishing orders: 6!. Step 2: 6! = 720. Step 3: In half of these permutations, No Hope finishes before Goldenrod. Step 4: In the other half, Goldenrod finishes before No Hope. Step 5: Therefore, number of permutations with No Hope ahead of Goldenrod = 720 / 2. Step 6: Compute 720 / 2 = 360. Step 7: Hence, 360 finishing orders satisfy the condition. Verification / Alternative check: As an alternative approach, you could treat N and G as a pair in each permutation and note that for any arrangement of the remaining 4 horses, there are exactly two ways to place N and G in the chosen positions: one where N is earlier and one where G is earlier. Thus, for every arrangement of the 4 other horses and fixed pair positions, half the arrangements have N ahead of G. This again shows that exactly half of the 720 total permutations satisfy the condition, confirming the answer 360. Why Other Options Are Wrong: 720: This is the total number of orders without any condition, counting both N before G and G before N. 120: This is 5!, not connected to the current situation where 6 horses participate. 640: A random number that does not emerge from any symmetric or factorial based reasoning for 6 horses. Only 360 correctly represents half of the total possible orders. Common Pitfalls: A common mistake is to attempt to count all permutations where N is before G directly without using symmetry, which can be complex and error prone. Another pitfall is to divide by an incorrect factor like 6 or 3 instead of 2, misinterpreting the relative positions. Remember that for two distinct labelled items in random permutations, each has an equal chance of being ahead of the other, giving a clean 1/2 probability and simplifying the count. Final Answer: The number of finishing orders in which No Hope finishes ahead of Goldenrod is 360.

Permutation and Combination Questions