Permutations & Combinations Practice Test (Chapter-8): 100 MCQs for Competitive Exams in India Quantitative AptitudeTest 1Test 2Test 3Test 4Test 50% Report a question What's wrong with this question?You cannot submit an empty report. Please add some details. Permutations & Combinations MCQs Test -420 MCQs20 Marks30 Minutes 1 / 20How many different 4-letter "words" can you form using the letters of the word "TENNIS"? 60 120 360 720 Explanation:To form 4-letter "words" using the letters of "TENNIS," you use permutations. However, there are repeated letters (2 'N's, 2 'E's, 2 'S's), so you divide by the factorial of the number of repetitions: 7! / (2! * 2! * 2!) = 120. 2 / 20There are three rooms in a motel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms? 7! / 1! 2! 3! 7! / 3! 7! / 3 7! 3 / 20In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position? 12 14 9 18 4 / 20In a group of 10 people, how many ways can you select a committee of 3 people and a chairperson from this group? 720 360 210 2,160 Explanation:To select a committee of 3 people and a chairperson from a group of 10 people, you're looking for permutations. First, choose 3 people from the group (C(10, 3)), and then choose one of them as the chairperson (3 choices). Multiply these choices: C(10, 3) * 3 = 120 * 3 = 360. However, this committee can be selected in any order, so you multiply by the number of permutations of 3 people: 360 * 3! = 2,160. 5 / 20How many different ways can you arrange the letters in the word "ELEPHANT" such that no two vowels are adjacent? 8,640 6,048 5,040 7,200 Explanation:To arrange the letters in "ELEPHANT" with no two vowels adjacent, you can treat the consonants (L, P, H, N, T) as a single entity. Then, you have 4 entities to arrange: (E, LPT, H, N, T). This can be done in 5! ways. However, within the LPT entity, there are 3! ways to arrange the consonants. For the E entity, there are 2! ways to arrange the vowels. So, the total arrangements are 5! * 3! * 2! = 6,048. 6 / 20In how many ways can six different rings be worn on four fingers of one hand ? 10 15 12 16 7 / 20The number of four–digit telephone numbers having at least one of their digits repeated is 4620 4960 4190 4680 8 / 20In a Sudoku puzzle, how many ways can you arrange the numbers 1 through 9 in the first row (from left to right) such that no two consecutive numbers are adjacent? 360 504 216 72 Explanation:To arrange the numbers 1 through 9 in the first row of a Sudoku puzzle without consecutive numbers, you need to count the permutations that satisfy this condition. You can treat the numbers 1 through 9 as "blocks" to arrange, and you have 3 "blocks" of odd numbers and 3 "blocks" of even numbers. There are 3! ways to arrange the odd-numbered blocks and 3! ways to arrange the even-numbered blocks. Therefore, the total number of arrangements is 3! * 3! = 36, and there are 14 ways to pick the order of odd and even blocks. So, the total number of arrangements is 36 * 14 = 504. 9 / 20In how many different ways can the letters of the word "COUNTRY" be arranged in such a way that the vowels always come together? 1440 5040 2880 720 10 / 20In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? 360 720 700 120 11 / 20From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ? 6400 3600 7600 7200 12 / 20There are 100 students in a college class of which 36 are boys studying statistics and 13 girls not studying statistics. If there are 55 girls in all, then the probability that a boy picked up at random is not studying statistics, is 3/5 4/5 2/5 1/5 13 / 20A box contains 5 red balls, 3 green balls, and 4 blue balls. How many ways can you select 3 balls from the box without looking if they are identical, and you don't care about the order? 220 120 330 660 Explanation:This is a combination problem without regard to order or the identical nature of the balls. You can use the combination formula to find the number of ways to choose 3 balls from the box: C(12, 3) = 220. 14 / 20A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is 346 280 196 140 15 / 20In a game, you have a deck of 40 cards, with 10 cards in each of 4 different suits. How many different 5-card hands can be drawn such that there is exactly one card from each suit? 5,040 11,880 20,160 8,640 Explanation:To form a hand with exactly one card from each suit, you can choose 1 card from each of the 4 suits (C(10, 1) for each suit), which gives you 10 * 10 * 10 * 10 = 10,000 combinations. However, you can choose any of the 5 cards to be the fifth card, so you multiply by 5, resulting in 10,000 * 5 = 50,000. However, there are 4 ways to choose which suit to have the additional card from, so you divide by 4, giving you 50,000 / 4 = 12,500. Finally, there are 5! ways to arrange the 5 cards, so you divide by 5!, resulting in 12,500 / 120 = 8,640 different hands. 16 / 20A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ? 186 190 200 196 17 / 20How many 4 digit numbers can be formed using the digits (1, 3, 4, 5, 7, 9) when repetition of digits is not allowed? 300 360 320 340 18 / 20In a chess tournament, there are 8 players, and each player plays against every other player exactly once. How many games are played in total? 32 24 16 28 Explanation:In a round-robin chess tournament with 8 players, each player plays against every other player once. To find the total number of games, you can use the combination formula, C(8, 2), which represents the number of ways to choose 2 players out of 8 to play against each other. Calculating it gives you 28 games. 19 / 20How many different 5-digit numbers can be formed using the digits 1, 2, 3, 4, and 5, with the condition that the number must be divisible by 5, and no digit can be repeated? 24 30 12 6 Explanation:To be divisible by 5, the last digit must be either 5 or 0. There are two options for the last digit (5 or 0). For the remaining four digits, there are 4! ways to arrange them without repetition. Therefore, the total number of such numbers is 2 * 4! = 12. 20 / 20How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? 2 3 4 6 Previous 1 2 3 4 5 6NextRelated Posts Average- Mean, Median & Mode (Chapter-1): 110 Important MCQs for Indian Competitive Exam Prep Simple & Compound Interest- Quiz (Chapter-2) : 100 Key MCQs to Competitive Exam Success Mastering Trigonometry for Quantitative Aptitude (Chapter-12): 100 MCQS to Ace Competitive Exams in IndiaLeave a Reply Cancel Reply Save my name, email, and website in this browser for the next time I comment. I have read and accepted the Privacy Policy * Δ
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