04 Topic wise Quantitative Aptitude Quiz with Solution for all Exam

Topic wise Quantitative Aptitude Quiz with Solution for all Exam

Quantitative Aptitude Quiz Topic: Permutation

1 >>Q1. In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate? ?

• (A) 144
• (B) 132
• (C) 156
• (D) 148

Solution Exp. Number of ways (4!)x(3!) = 144.

2 >>Q2. In how many ways 4 boys and 4 girls can be seated in a row so that boys and girls are alternate? ?

• (A) 2152
• (B) 1146
• (C) 1152
• (D) 1278

Solution Exp. Either a boy or a girl can be first.Hence number of ways = 2x(4!)x(4!) = 1152.

3 >>Q3. There are 5 boys and 3 girls. In how many ways can they be seated in a row so that all the three girls do not sit together? ?

• (A) 16000
• (B) 18000
• (C) 19000
• (D) 36000

Solution Exp. Number of ways all three girls sit together = 3! x 6! = 4320.Total number of ways they can be seated = 8! = 40320.Number of ways they can be seated so that all three girls do not sit together = 40320 - 4320 = 36000.

4 >>Q4. How many different words can be formed with the letters of the word 'PENCIL' when vowels occupy even places? ?

• (A) 144
• (B) 248
• (C) 288
• (D) 72

Solution Exp. Number of ways 2 vowels can be placed in 3 even places = 2 x (3!/2!) = 6.Number of ways the 4 consonants can be placed in remaining places = 4! = 24.Total number of ways = 6x24 = 144.

Download 5 >>Q5. In how many ways can the letters of the word 'DIRECTOR' be arranged so that the three vowels are never together? ?

• (A) 9000
• (B) 18000
• (C) 16000
• (D) 19000

Solution Exp. Number of ways it can arranged such that three vowels occur together = (3x(8-3+1)!)/2! = 2160 (R occurs twice, hence dividing by 2!).Number of ways all letters can be arranged = 8!/2! = 20160.Number of ways the letter can be arranged such that the three vowels do not occur together = 20160 - 2160 = 18000.

6 >>Q6. How many different letter arrangement can be made from the letters of the word 'RECOVER'? ?

• (A) 1260
• (B) 1560
• (C) 2360
• (D) 1256

Solution Exp. Number of letter arrangements = (7!)/(2! x 2!) = 1260 (in the 7 letters, E and R occur twice hence dividing twice by 2!).

7 >>Q7. From four officers and eight jawans in how many ways can be six chosen include at least one officer? ?

• (A) 796
• (B) 996
• (C) 556
• (D) 896

Solution Exp. Number of ways six can be chosen from the 12 = 12C6 = 924.Number of ways only jawans can be chosen = 8C6 = 28.Number of ways six can be chosen with atleast one officer = 924 - 28 = 896.

8 >>Q8. The number of straight lines can be formed out of 10 points of which 7 are collinear? ?

• (A) 65
• (B) 45
• (C) 25
• (D) 35

Solution Exp. Number of lines formed using 10 points = 10C2 = 45.Number of lines formed by 7 points = 7C2 = 21.Total number of lines = 45 - 21 + 1(collinear line) = 25.

9 >>Q10. Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. How many persons in the room? ?

• (A) 12
• (B) 11
• (C) 13
• (D) 14

Solution Exp. Let total number of people = X.Number of handshakes = XC2 = 66.(X)(X-1)/2 = 66.X = 12.

10 >>Q12. A polygon has 44 diagonals the number of its sides is- ?

• (A) 11
• (B) 13
• (C) 9
• (D) 17

Solution Exp. Let number of sides = N.Number of diagonals = N(N-3)/2 = 44.N = 11.

11 >>Q13. The number of different permutations of the word 'BANANA' is- ?

• (A) 20
• (B) 30
• (C) 60
• (D) 120

Solution Exp. Number permutations = 6! / (3! x 2!) = 60 (A occurs thrice, N occurs twice, hence dividing by 3! and 2!).

12 >>Q15. If nPr = 120 nCr, then r is equa1 to- ?

• (A) 6
• (B) 7
• (C) 3
• (D) 5

Solution Exp. n! / (n-r)! = 120 x n! / (r! x (n-r)!).1 = 120 /r!.r! = 120, r = 5.

13 >>Q16. The total number of permutations of four letters that can be made out of the letters of the word 'EXAMINATION' is- ?

• (A) 2454
• (B) 3454
• (C) 2001
• (D) None of these

Solution Exp. There are 8 letters with A, N and I repeating twice.Number of ways a four letter word can be formed with all different letters = (8!/4!) = 1680.Number of four letter words with one repeated letter = 3C1 x 4C2 x 7P2 = 756 (Choose 1 of the 3 repeatable letters; choose 2 of the 4 slots in permutation for the letter; fill the remaining 2 slots from the 7 remaining letters).Number of four letter words with two repeated letters = 3C2 x 4C2 x 2C2 = 18 (Choose 2 of the 3 repeatable letters; choose 2 of the 4 slots for the first letter; choose the 2 remaining slots for the second letter).Total = 1680 + 756 + 18 = 2454.

14 >>Q18. In how many different ways can the letters of the word 'PADDLED' be arranged? ?

• (A) 740
• (B) 840
• (C) 640
• (D) 540

Solution Exp. Number of ways = 7! / 3! = 840 (D repeats thrice, hence dividing by 3!).

Download 15 >>Q19. The number of triangles that can be formed by choosing the vertices from a set of 12 points, 7 of which lie on the same straight line, is- ?

• (A) 185
• (B) 285
• (C) 555
• (D) 625

Solution Exp. Number of triangles that can be formed using 12 points = 12C3 = 220.Number of triangles that can be formed using 7 points = 7C3 = 35.Hence, number of triangles = 220 - 35 = 185.

16 >>Q20. If S = {2, 3, 4, 5, 7, 9}, then the number of different three-digit numbers (with all distinct digits) less than 400 that can be formed from S is- ?

• (A) 21
• (B) 25
• (C) 44
• (D) 40

Solution Exp. First digit is either 2 or 3, that is 2 ways.Next two digits are chosen from the remaining 5 digits = 5P2 = 20.Total number of ways = 2x20 = 40.

17 >>Q22. How many words can be formed out of the letters of the word 'VELOCITY', so that vowels occupy the even place? ?

• (A) 720
• (B) 480
• (C) 17280
• (D) 2880

Solution Exp. Number of ways 3 vowels can occupy 4 even places = 4P3 = 24.Number of ways remaining 5 letters occupy the space = 5! = 120.Total number of ways = 24 x 120 = 2880.

18 >>Q26. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can the balls be place so that no box remains empty? ?

• (A) 160
• (B) 150
• (C) 140
• (D) 190

Solution Exp. If one box gets one ball and the other two get two balls each, number of ways = 5 x 3C1 x 4C2 = 90.If one box gets three balls and the other two get one ball each, number of ways = 5 x 5C3 x 2C1 = 60.Total = 90 + 60 = 150.

19 >>Q28. 12 persons are to be arranged to a round table. If two particular person among them are not to be side by side the total number of arrangement is- ?

• (A) 9( 10!)
• (B) 2 (10!)
• (C) 45 (8!)
• (D) 10!

Solution Exp. Number of ways 12 person can be arranged in a round table = (12-1)! = 11!.Number of ways two particular people are placed side by side = 2(11-1)! = 2 x 10! (considering those two people as one unit, we have 11 places).Number of ways they do not sit together = 11! - 2 x 10! = 10! (11 - 2) = 10!(9).

20 >>Q29. The number of odd integers between 1000 and 9999 with no digit repeated is- ?

• (A) 3333
• (B) 2120
• (C) 2240
• (D) 3331

Solution Exp. We have 5 odd numbers, placing it in the last place gives us 5 possibilities.The first digit can be anything from 1 to 9 excluding the odd number we placed at the end, hence we have 9-1 = 8 ways.The second digit can be anything from 0 to 9 excluding the odd number and the first digit = 8 ways.The third digit can be from 0 to 9 excluding three digits = 7.Total number of ways = 5x8x8x7 = 2240.