Permutation And Combination – Aptitude Questions and Answers | Latest Online Permutation And Combination MCQ Aptitude Test
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Permutation And Combination – Aptitude Questions and Answers | Online Permutation And Combination MCQ Aptitude Test
Quiz Name
Permutation And Combination
Category
Online Aptitude Test
Number of Questions
15
Time
30 Minutes
Exam Type
MCQ (Multiple Choice Questions)
1. Five people will be chosen from a group of seven men and six women to create a committee, with at least three males on the committee. How many different ways can it be done? a. 564 b. 645 c. 735 d. 756
Answer: D
Explanation: We might have 3 men and 2 women, 4 men and 1 woman, or 5 men and 1 woman (5 men only 7 x 6 x 5/3 x 2 x 1 2 x 1 x 6 x 5/2 x1 + (7C3 x 6C1) + (7C3 x 6C2) + (7C4 x 6C1) + (7C5) = 7 x 6 x 5/3 x 2 x 1 2 x 1 x 6 x 5/2 x1 + (7C3 x 6C1) + (7C3 x 6C1) + (7C3 x 6 (7C2)= 525 + 7 x 6 x 5 x 6/3 x 2 x 1 2 x 1 + 7 x 6/2 x 1 = (756 + 525 + 210 + 21).
2. How many different ways can the letters in the word ‘LEADING’ be arranged so that the vowels are always together? a. 360 b. 480 c. 720 d. 5040
Answer: C
Explanation: There are seven letters in the word ‘LEADING.’When the vowels EAI are constantly together, they are thought to make a single letter.The letters LNDG must then be arranged (EAI).Now you can arrange 5 (4 + 1 = 5) letters in 5! = 120 different ways.The vowels (EAI) can be placed in 3! = 6 different ways.The required number of ways is 720 (120 x 6).
3. How many different ways can the letters in the word “CORPORATION” be arranged so that the vowels always meet? a. 810 b. 1440 c. 2880 d. 50400
Answer: D
Explanation: The vowels OOAIO are treated as one letter in the word ‘CORPORATION.’As a result, we have CRPRTN (OOAIO).This has seven letters (six plus one), with R appearing twice and the rest being unique.The number of different ways to arrange these letters is 7!/2! = 2520.Now, 5 vowels may be arranged, with O appearing three times and the others being differen in 5!/3! = 20 different ways.The required number of ways is 50400 (2520 x 20).
4. How many words may be created with three consonants and two vowels from a set of seven consonants and four vowels? a. 210 b. 1050 c. 25200 d. 21400
Answer: C
Explanation: There are three consonants out of seven to choose from, and (2 vowels out of 4).7 x 6 x 5/3 x 2 x 1 2 x 1 x 4 x 3/2 x 1 = (7C3 x 4C2) = 7 x 6 x 5/3 x 2 x 1 2 x 1 x 4 x 3/2 x 1equals 210,There are 210 groups, each with three consonants and two vowels.There are five letters in each group.The number of different ways to arrange five letters is 5! = 5 x 4 x 3 x 2 x 1 = 120.Number of methods required = (210 x 120) = 25200.
5. How many different ways can you arrange the letters in the word ‘LEADER’? a. 72 b. 144 c. 360 d. 720
Answer: C
Explanation: The word ‘LEADER’ is made up of six letters: 1L, 2E, 1A, 1D, and 1R.6!/ (1!)(2!)(1!)(1!)(1!)(1!)=360.
6. Four youngsters will be chosen from a group of six males and four girls. How many different ways can they be chosen to ensure that at least one boy is present? a. 159 b. 194 c. 205 d. 209
Answer: D
Explanation: We might have one boy and three girls, or two boys and two girls, or three boys and one female (4 boys).6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) = (6 x 4) +6 x 5/2 x 1 x4 x 3/2 x 1 +6 x 5 x 4 x 4/ 3 x 2 x 1 2 x 1+6 =209.
7. How many three-digit numbers can be made using the digits 2, 3, 5, 6, 7, and 9, all of which are divisible by five and have no repeats? a. 5 b. 10 c. 15 d. 20
Answer: D
Explanation: We must have 5 in the unit position because each required value is divisible by 5. So there’s only one way to go about it.Any of the remaining 5 numbers can now occupy the tens slot (2, 3, 6, 7, 9). There are five different methods to fill the tens position.Any of the remaining four digits can now be used to fill the hundreds position. So there are four options for filling it.(1 x 5 x 4) = 20 is the required number of numbers.
8. How many different ways can a committee of 5 men and 6 women be created from 8 men and 10 women? a. 266 b. 5040 c. 11760 d. 86400
Answer: C
Explanation: 8 x 7 x 6 x 10 x 9 x 8 x 7 3 x 2 x 1 4 x 3 x 2 x 1 = 11760 = (8C5 x 10C6) = (8C3 x 10C4) = 8 x 7 x 6 x 10 x 9 x 8 x 7 3 x 2 x 1 4 x 3 x 2 x 1=11760
9. There are two white balls, three black balls, and four red balls in a box. If at least one black ball is to be included in the draw, how many ways may three balls be chosen from the box? a. 32 b. 48 c. 64 d. 96
Answer: C
Explanation: We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black and 1 non-black) (3 black).3 x 6 x 5/2 x 1 + 3 x 2/2 x 1 x 6 + 1 = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) = 3 x 6 x 5/2 x 1 + 3 x 2/2 x 1 x 6 + 1,(45 + 18 + 1) equals 64.
10. How many different ways can the letters in the word ‘DETAIL’ be arranged so that the vowels only appear in odd positions? a. 32 b.48 c. 36 d. 60
Answer: C
Explanation: There are six letters in the given word, three vowels and three consonants.Let’s label these areas as follows:(1) (2) (3) (4) (5) (6) (7) (8) (8) (8) (8) (8) (6),Three vowels can now be put in any of the three spaces marked 1, 3, and 5.The number of different ways to arrange the vowels is 3P3 = 3! = 6.The three consonants can also be placed in any of the other three slots.The number of different ways these configurations may be made is 3P3 = 3! = 6.The total number of ways is 36 (6 × 6).
11. Out of a total of 7 men and 3 women, how many ways can a group of 5 men and 2 women be formed? a. 63 b. 90 c. 126 d. 45
Answer: A
Explanation: 7 x 6/2 x 1 x 3 = 63 = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6/2 x 1 x 3 =63
12. If letter repetition is not permitted, how many four-letter words with or without meaning can be created from the letters in the word ‘LOGARITHMS’? a. 40 b. 400 c. 5040 d. 2520
Answer: C
Explanation: ‘LOGARITHMS’ is made up of ten distinct letters.The required number of words is equal to the number of 10 letter arrangements taken four at a time. = 5040 = 10P4 = (10 x 9 x 8 x 7).
13. How many different ways can the letters in the word “MATHEMATICS” be arranged so that the vowels always meet? a. 10080 b. 4989600 c. 120960 d. None of these
Answer: C
Explanation: The vowels AEAI are treated as one letter in the word ‘MATHEMATICS.’As a result, we have MTHMTCS (AEAI).Now we must arrange eight letters, with M appearing twice, T appearing twice, and the remainder being unique.The number of different ways to arrange these letters is 8!/ (2!)(2!) = 10080.AEAI now contains four letters, with A appearing twice and the rest being unique.The number of different ways to arrange these letters is 4!/2! = 12.(10080 x 12) = 120960 is the required number of words.
14. How many different ways can you arrange the letters in the word “OPTICAL” so that the vowels always come together? a. 120 b. 720 c. 4320 d. 2160
Answer: B
Explanation: The word ‘OPTICAL’ has seven different letters in it.When the vowels OIA are constantly together, they are thought to make a single letter.The letters PTCL must then be arranged (OIA).5 letters can now be placed in 5! = 120 different ways.The vowels (OIA) can be placed in 3! = 6 different ways.The required number of ways is 720 (120 x 6).
