Permutation Quiz

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.

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!).

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 >>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).

19 >>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.

20 >>Q30. The number of different messages that can be represented by three 0's and two I's is- ?

• (A) 10
• (B) 6
• (C) 9
• (D) 8

Solution Exp. Number of messages = 5! / (3! x 2!) = 10 (total number of units = 3+2 = 5, 3 repeating 0's and 2 repeating I's).