Permutations & Combinations Practice Test (Chapter-8): 100 MCQs for Competitive Exams in India September 16, 2023 Quantitative AptitudeTest 1Test 2Test 3Test 4Test 5 Report a question What's wrong with this question?You cannot submit an empty report. Please add some details. 0% Permutations & Combinations Mock Quiz-320 MCQs20 Marks30 Minutes 1 / 20In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together? 8 x 8! 8 x 9! 7 x 9! 9 x 8! 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! 2 / 20How many different ways can you arrange the letters in the word "BANANA"? 120 240 720 60 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. 3 / 20If you have 5 different books and you want to arrange them on a shelf, how many different arrangements are possible? 60 120 25 15 Explanation:To arrange 5 different books on a shelf, you use permutations. The number of permutations is 5!. 4 / 20How many different pairs of socks can you make from a drawer containing 5 pairs of red socks, 4 pairs of blue socks, and 3 pairs of green socks? 132 30 60 12 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. 5 / 20The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by 7! x 5! 6! x 5! 5! x 4! 30 6 / 20How many different arrangements can be made using the letters of the word "COMPUTER" where no two vowels are adjacent? 2,520 4,320 1,680 3,360 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. 7 / 20How many different 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5, with no digit repeated? 120 60 20 30 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. 8 / 20A company has 7 employees, and they want to form a project team with 4 members. If the team must include exactly 2 software engineers and 1 manager, how many different teams can be formed? 42 84 21 35 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. 9 / 20How many 3-digit numbers can be formed from the digit 2,3,5,6,7 and 9, which are divisible by 5 and none of the digits is repeated? 10 5 15 20 10 / 20A box contains 6 red balls and 4 blue balls. How many different ways can you select 3 balls from the box without replacement if at least one ball must be red? 36 40 84 128 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. 11 / 20A committee of 5 students is to be formed from a class of 20 students. How many different committees can be formed if the committee can consist of at most 2 girls? 18,564 20,125 14,560 38,760 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. 12 / 20In a coding competition, there are 6 problems to solve. If a participant can choose to solve any number of problems, how many different ways can they choose a set of problems to solve? 36 720 2^6 64 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. 13 / 20How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines? 315 311 339 312 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 14 / 20How many ways can you arrange the letters in the word "SUCCESS" such that no two 'S's are adjacent? 1,260 3,150 5,040 2,520 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. 15 / 20In how many different ways can you arrange the digits 1, 2, 3, 4, 5, 6, 7, and 8 to form a four-digit number with no repetition? 1008 1680 40320 5040 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. 16 / 20In a deck of 52 playing cards, how many different five-card poker hands can be formed that contain exactly two pairs (e.g., two Aces, two Kings, and one other card)? 1,440 108 360 2,016 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. 17 / 20In a deck of 52 playing cards, how many different five-card poker hands can be formed that contain a full house (three of a kind and a pair)? 532 156 744 312 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. 18 / 20A group of 10 people is going on a road trip, and they want to sit in a row of 10 seats in a van. How many different seating arrangements are possible? 90,000 10,000 3,628,800 100 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. 19 / 20In a deck of 52 playing cards, how many different ways can you select 3 cards, allowing for duplicates (repetition) in the selection? 22,100,600 23,416,600 25,852,600 22,100 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. 20 / 20How many different seven-letter words can be formed using the letters of the word "REFRESH"? 25200 720 2520 5040 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!. 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Speed, Time & Distance Practice MCQs (Chapter-7): 100 Important Questions for Quantum Aptitude Preparation