Permutations & Combinations Practice Test (Chapter-8): 100 MCQs for Competitive Exams in India

Explanation:

To arrange the letters in "SUCCESS" such that no two 'S's are adjacent, treat the 'S's as separators for the other letters ('C,' 'U,' 'C,' 'E'). You have 5 entities to arrange: 'S,' 'C,' 'U,' 'C,' 'E.' This can be done in 5! ways. However, within the 'C,' 'U,' 'C,' entity, there are 2! ways to arrange the 'C's. So, the total arrangements are 5! / 2! = 1,260.

Explanation:

When you allow duplicates (repetition), you can choose any of the 52 cards for each of the 3 selections, leading to a total of 52^3 = 25,852,600 different ways.

Explanation:

To arrange the digits with no repetition, you have 8 choices for the first digit, 7 choices for the second digit, 6 choices for the third digit, and 5 choices for the fourth digit. Multiply these choices: 8 * 7 * 6 * 5 = 1680.

Explanation:

To form seven-letter words using the letters of "REFRESH," you use permutations. There are 7 distinct letters in the word, so the number of permutations is 7!.

Explanation:

To form a team with 2 software engineers, 1 manager, and 1 other member, you can calculate it as follows: C(2, 2) * C(1, 1) * C(4, 1) = 1 * 1 * 4 = 4. However, you can choose any 2 out of the 7 employees to be software engineers, so you multiply by C(7, 2) = 21. The total number of teams is 4 * 21 = 84.

Explanation:

To form a committee of 5 students with at most 2 girls, you can calculate the total number of committees minus the committees with no girls and the committees with only one girl. Total committees = C(20, 5). Committees with no girls = C(15, 5). Committees with only one girl = C(5, 1) * C(15, 4). Subtracting these from the total gives you C(20, 5) - C(15, 5) - [C(5, 1) * C(15, 4)] = 38,760.

Explanation:

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! x 2!
= 9!(10 - 2)
= 8 x 9!

Explanation:

To find the number of ways to select 3 balls with at least one red ball, you can calculate the total number of ways to choose 3 balls (C(10, 3)) and subtract the number of ways to choose 3 blue balls (C(4, 3)). So, C(10, 3) - C(4, 3) = 120 - 4 = 84 ways.

Explanation:

To form a 3-digit number using these digits with no repetition, you have 5 choices for the first digit, 4 choices for the second digit (as one digit has already been used), and 3 choices for the third digit (two digits have already been used). Multiply these choices: 5 * 4 * 3 = 60. So, there are 120 different 3-digit numbers.

Explanation:

For each problem, the participant has two choices: to solve it or not to solve it. Since there are 6 problems, the total number of ways they can choose a set of problems to solve is 2^6, which is 64.

Explanation:

To form a full house, you first choose a rank for the three of a kind (C(13, 1) = 13 options), then choose 3 suits for those cards (C(4, 3) = 4 options). Then, you choose a rank for the pair (C(12, 1) = 12 options) and choose 2 suits for those cards (C(4, 2) = 6 options). Multiply these choices: 13 * 4 * 12 * 6 = 3,744. However, there are 6 ways to arrange the three of a kind and the pair, so you divide by 6 to get 3,744 / 6 = 624. Finally, note that there are 2 possible orders for a full house (e.g., AAAKK and KKKAa), so you divide by 2, resulting in 624 / 2 = 312.

Explanation:

To find the number of different pairs of socks, you need to count the combinations of choosing 2 socks from each color group and then summing them up: C(5, 2) + C(4, 2) + C(3, 2) = 10 + 6 + 3 = 19 pairs. However, these pairs can be of any color, so you need to multiply by 3 (the number of color options) to get 19 * 3 = 57 pairs.

Explanation:

To arrange the letters in "COMPUTER" with no two vowels adjacent, consider the consonants as a single entity (C, M, P, T, R) and the vowels as another entity (O, U, E). You now have 2 entities to arrange: (CMPT, OUR). This can be done in 5! * 4! = 2,520 ways.

Explanation:

To arrange 5 different books on a shelf, you use permutations. The number of permutations is 5!.

Explanation:

To form a poker hand with two pairs, choose two distinct ranks for the two pairs (C(13, 2) ways), choose one of the four suits for each pair (4 choices for each pair), and choose one card from the remaining 11 ranks for the fifth card (C(11, 1) ways). Multiply these choices: C(13, 2) * (4^2) * C(11, 1) = 1,440.

Explanation:

To find the number of ways to arrange the letters in the word "BANANA," you use permutations because the order of the letters matters. However, there are repeated letters (3 'A's and 2 'N's), so you divide by the factorial of the number of repetitions. The calculation is 6! / (3! * 2!) = 120.

Explanation:

This is a permutation problem. There are 10 people, and you want to arrange them in a row of 10 seats. This can be done in 10! (10 factorial) ways, which is equal to 3,628,800.

Explanation:

Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315