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 and Combinations Mock Test-220 MCQs20 Marks30 Minutes 1 / 20A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt? 12 21 9 108 Explanation:The boy can select one trouser in nine ways. The boy can select one shirt in 12 ways. The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways. 2 / 20A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation? 20 24 16 28 Explanation:There are three ladies and five gentlemen and a committee of 5 members to be formed) Number of ways such that two ladies are always included in the committee = 6C3 = (6 * 5 * 4) / 6 = 20 3 / 20In a city with 7 bridges, each bridge connects a different pair of islands. If you want to take a tour, starting and ending at the same island, without crossing any bridge more than once, how many different tours are possible? 6 14 20 7 Explanation:This problem is a variation of the famous "Seven Bridges of Konigsberg" problem. To find the number of different tours, you need to consider Eulerian circuits. In this case, there are 14 different tours that satisfy the conditions. 4 / 20How many ways can you arrange the letters in the word "MISSISSIPPI"? 34650 2520 36720 31920 Explanation:This is a permutation problem with repeated letters. There are 11 total letters, with 4 'I's, 4 'S's, 2 'P's, and 1 'M'. The number of permutations is 11! / (4! * 4! * 2!). 5 / 20In a chess tournament, there are 10 players. How many different ways can the top 3 players be awarded gold, silver, and bronze medals? 30 120 720 210 Explanation:To award the top 3 players medals, you're looking for permutations. There are 10 choices for the gold medalist, 9 choices for the silver medalist (as one player has already received gold), and 8 choices for the bronze medalist (with two players already awarded medals). So, the total number of permutations is 10 * 9 * 8 = 720. However, since the order of awarding medals doesn't matter, you need to divide by 3! (the number of ways to arrange 3 medals), so the final count is 720 / 6 = 210. 6 / 20How many ways can you arrange the letters in the word "CHAIR"? 120 6 720 24 Explanation:To find the number of ways to arrange the letters in the word "CHAIR," you use permutations because the order of the letters matters. There are 5 distinct letters in the word, so the number of permutations is 5!. 7 / 20A box contains three white balls, four black balls and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is 600 500 100 400 Explanation:The required number of ways a) 1 black and 2 others = 4C1.6C2 = 4 x 15 = 60 b) 2 black and 1 other = 4C2.6C1 = 6 x 6 = 36 c) All the three black = 4C3 = 4 Total =60 + 36 + 4 = 100 8 / 20How many ways can you arrange the letters in the word "APPLE"? 60 5 720 120 Explanation:To find the number of ways to arrange the letters in the word "APPLE," you use permutations because the order of the letters matters. There are 5 letters in total, with two 'P's, so the number of permutations is 5! / 2! = 120. 9 / 20A polygon has 44 diagonals. What is the number of its sides? 12 11 9 10 Explanation:The number of diagonals is given by the formula nC2 - n 10 / 20You have 8 different colored marbles, and you want to pick 3 of them. How many different combinations of 3 marbles can you choose? 56 336 120 8 Explanation:This is a combination problem. To find the number of different combinations of 3 marbles out of 8, you can use the combination formula, which is C(8, 3) = 8! / (3!(8-3)!). Calculating it gives you 56. 11 / 20A pizza menu offers 5 different toppings, and you can choose up to 3 toppings for your pizz 30 How many different pizzas can you create? 20 10 26 Explanation:This is a combination problem because the order of topping selection doesn't matter. You can choose 0, 1, 2, or 3 toppings from the 5 available. To find the total number of different pizzas, sum the combinations for each option: C(5, 0) + C(5, 1) + C(5, 2) + C(5, 3) = 1 + 5 + 10 + 10 = 26. So, you can create 26 different pizzas. 12 / 20In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty? 20 45 36 14 Explanation:This is nothing but the number of ways of having a, b, c such that a + b + c = 11, where a, b, c are natural numbers. By having them to be natural numbers, we ensure that no box can be empty. (no zeroes)- 10C2 = 45 ways 13 / 20In a class of 20 students, how many different committees of 4 students can be formed? 80 4845 5 1140 Explanation:This is a combination problem because the order of selection doesn't matter. You can use the combination formula, which is C(20, 4) = 20! / (4!(20-4)!), to find the number of ways to choose the committee. Calculating it gives you 4845. 14 / 20How many different ways can you distribute 12 identical candies among 4 children, such that each child receives at least one candy? 120 360 140 126 Explanation:This problem is a combination with repetition. You can use the "stars and bars" method to distribute the candies. In this case, you need to distribute 12 identical candies into 4 distinct bins (children). Using the formula C(12 + 4 - 1, 4 - 1), you get C(15, 3) = 126 different ways to distribute the candies among the children. 15 / 20How many different four-digit numbers can be formed using the digits 1, 2, 3, 4, 5, and 6, without repetition, and where the number is divisible by 4? 48 120 144 72 Explanation:To find four-digit numbers divisible by 4, the last two digits must form a number divisible by 4. There are six choices for the last digit and five choices for the third digit (without repetition). So, there are 6 * 5 = 30 possibilities for the last two digits. For the first two digits, you can use any of the remaining four digits, resulting in 4 * 30 = 120 combinations. However, half of these will end in an even number and not be divisible by 4, so the final count is 120 / 2 = 72. 16 / 20In how many ways can you choose a president, vice-president, and treasurer from a group of 10 people? 10 30 720 120 Explanation:To choose a president, vice-president, and treasurer, you're looking for permutations. There are 10 choices for the president, 9 choices for the vice-president (since one person has already been chosen as president), and 8 choices for the treasurer (as two people have already been chosen). So, the total number of permutations is 10 * 9 * 8 = 720. 17 / 20A restaurant offers a menu with 10 appetizers, 12 main courses, and 8 desserts. How many different three-course meals (one appetizer, one main course, and one dessert) can you create? 2,880 9,600 1,920 6,400 Explanation:To create a three-course meal, you can choose one appetizer from 10 options, one main course from 12 options, and one dessert from 8 options. The total number of different three-course meals is 10 * 12 * 8 = 9,600. 18 / 20In a deck of 52 playing cards, how many different five-card poker hands can be formed that contain exactly two red cards and three black cards? 14880 11760 9240 6840 Explanation:To form a five-card poker hand with two red cards (diamonds and hearts) and three black cards (clubs and spades), you can choose 2 red cards out of 26 (C(26, 2)) and 3 black cards out of 26 (C(26, 3)), and then multiply these two combinations to find the total number of such hands: C(26, 2) * C(26, 3) = 6840. 19 / 20In how many ways can you arrange the letters of the word "ALGORITHM" such that no two 'G's are adjacent? 4,320 5,040 6,480 7,200 Explanation:To arrange the letters of "ALGORITHM" such that no two 'G's are adjacent, treat the 'GG' combination as a single entity. Then, you have 7 entities ('A,' 'L,' 'O,' 'R,' 'I,' 'T,' and 'GG') to arrange. This can be done in 7! ways. However, within the 'GG' entity, there are 2! ways to arrange the two 'G's. So, the total arrangements are 7! / 2! = 6,480. 20 / 20How many ways can you arrange the letters in the word "MATH"? 20 24 4 12 Explanation:To find the number of ways to arrange the letters in the word "MATH," you use permutations because the order of the letters matters. There are 4 distinct letters in the word, so the number of permutations is 4!. Previous 1 2 3 4 5 6NextRelated Posts Profit and Loss MCQs (Chapter-4): 100 Essential Questions for Competitive Exams in India 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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