Permutations and Combinations Practice Questions

Correct Answer: B

This is a permutation problem because the order of the winners matters (gold is different from silver). We need to find the number of permutations of 8 contestants taken 3 at a time, which is denoted as 8P3. The formula for nPr is n! / (n-r)!. 8P3 = 8! / (8-3)! = 8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336. Therefore, there are 336 different ways to award the medals.

Correct Answer: C

This is a combination problem because the order in which the books are chosen does not matter. We need to find the number of combinations of 10 books taken 4 at a time, which is denoted as 10C4. The formula for nCr is n! / [r!(n-r)!]. 10C4 = 10! / [4!(10-4)!] = 10! / (4!6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 10 × 3 × 7 = 210. Therefore, there are 210 different combinations of books.

Correct Answer: D

The word 'APPLE' has 5 letters in total. The letter 'P' is repeated 2 times. To find the number of distinct arrangements, we use the formula for permutations with identical items: n! / (p1! * p2! * ...), where n is the total number of letters and p1, p2, ... are the frequencies of each repeated letter. Here, n = 5 and the letter 'P' is repeated 2 times. Number of arrangements = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 120 / 2 = 60. Therefore, there are 60 distinct arrangements.

Correct Answer: A

The word 'CRYSTAL' has 7 distinct letters: C, R, Y, S, T, A, L. The vowels are A. The consonants are C, R, Y, S, T, L. Wait, the question implies multiple vowels. Let's re-evaluate. The word is 'CRYSTAL'. The only vowel is 'A'. Let's change the word to 'FORMULA'. The word 'FORMULA' has 7 letters. The vowels are O, U, A (3 vowels). The consonants are F, R, M, L (4 consonants). To keep the vowels together, we treat the group of vowels (O, U, A) as a single unit. Now we have 4 consonants + 1 group of vowels = 5 units to arrange. These 5 units can be arranged in 5! ways. The 3 vowels within their group can be arranged in 3! ways. Total arrangements = 5! * 3! = 120 * 6 = 720.

Correct Answer: C

The problem requires forming a committee of 5 members with a specific composition: 3 men and 2 women. We need to select 3 men from 6 available men, and 2 women from 4 available women. The number of ways to select 3 men from 6 is given by the combination formula C(6, 3). The number of ways to select 2 women from 4 is given by C(4, 2). C(6, 3) = 6! / (3! * (6-3)!) = (6 * 5 * 4) / (3 * 2 * 1) = 20. C(4, 2) = 4! / (2! * (4-2)!) = (4 * 3) / (2 * 1) = 6. Since both selections must happen, we multiply the results: Total ways = 20 * 6 = 120. Wait, let me re-calculate C(6,3) is 20, C(4,2) is 6. 20*6=120. Let me check my options. Let's make the correct option 120. Let's change the question slightly. Committee of 4 members with exactly 2 men. Select 2 men from 6: C(6,2) = 6*5/2 = 15. Select 2 women from 4: C(4,2) = 4*3/2 = 6. Total ways = 15 * 6 = 90. Let's use this. A committee of 4 members is to be formed from a group of 6 men and 4 women. In how many ways can this be done if the committee must consist of exactly 2 men?

Correct Answer: C

We need to form a 4-digit number using the digits {1, 2, 3, 4, 5} without repetition. For a number to be divisible by 2, its last digit (the units place) must be an even number. The available even digits are 2 and 4. So, there are 2 choices for the units place. Once the units place is filled, we have 4 remaining digits for the other 3 places. The thousands place can be filled in 4 ways. The hundreds place can be filled in 3 ways. The tens place can be filled in 2 ways. Total ways = (Choices for thousands) * (Choices for hundreds) * (Choices for tens) * (Choices for units) = 4 * 3 * 2 * 2 = 48. Alternatively, fix the units place first. Case 1: Units place is 2 (1 way). The remaining 3 places must be filled from the remaining 4 digits {1, 3, 4, 5}. This can be done in P(4, 3) = 4!/(4-3)! = 24 ways. Case 2: Units place is 4 (1 way). The remaining 3 places must be filled from the remaining 4 digits {1, 2, 3, 5}. This can be done in P(4, 3) = 24 ways. Total ways = 24 + 24 = 48.

Correct Answer: C

This problem can be solved using the complementary counting principle. First, we find the total number of ways to arrange the 6 books without any restrictions. Then, we find the number of ways to arrange the books such that the two particular books are always together. Finally, we subtract the second result from the first. Total arrangements of 6 different books = 6! = 720. Now, find arrangements where two particular books (say B1 and B2) are always together. Treat (B1, B2) as a single unit. Now we have 5 units to arrange (the block of two books and the other 4 books). These can be arranged in 5! ways. The two books within the block can be arranged in 2! ways (B1B2 or B2B1). So, arrangements where they are together = 5! * 2! = 120 * 2 = 240. The number of arrangements where the two books are never together = (Total arrangements) - (Arrangements where they are together) = 720 - 240 = 480.