# RRB NTPC - Permutations Combinations

1.

In how many different number of ways 4 boys and 2 girls can sit on a bench?

A.) 620
B.) 640
C.) 720
D.) None of these

npn = n!
6p6
= 6 × 5 × 4 × 3 × 2 × 1 = 720

2.

In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons?

A.) 230
B.) 203
C.) 220
D.) 210

7p3 = 7 × 6 × 5 = 210

3.

In how many different number of ways 4 boys and 3 girls can sit on a bench such that girls always sit together.

A.) 720
B.) 5040
C.) 4320
D.) None of these

4.

In how many different ways can the letters of the word "CLAIM" be rearrangement?

A.) 120
B.) 125
C.) 130
D.) None of these

The total number of arrangements is
5P5  = 5! = 120

5.

If the letters of the word PLACE are arranged taken all at a time, find how many do not start with AE.

A.) 142
B.) 141
C.) 114
D.) None of these

Total no'of arrangements 5P5  = 5! = 120
no'of arrangements which do not start with AE = 120 - 6 = 114.

6.

How many arrangements of the letters of the word BEGIN can be made, without changing the place of the vowels in the word?

A.) 7 ways
B.) 6 ways
C.) 5 ways
D.) 2 ways

E,I fixed. Consonants can be arrangements in 3P3 = 3! = 6 ways

7.

If all the numbers 2, 3, 4, 5, 6, 7, 8 are arranged, find the number of arrangements in which 2, 3, 4, are together?

A.) 720
B.) 620
C.) 700
D.) None of these

If (2 3 4) is one.
we must arrange (2 3 4), 5, 6, 7, 8 in
5P5 = 5! = 120 ways
2, 3, 4 can be arranged in 3P3 = 3! = 6
120 × 6 = 720.

8.

Find 10P6

A.) 150200
B.) 151200
C.) 152200
D.) None of these

10P6 = 10!/4! = 10 × 9 × 8 × 7 × 6 × 5
= 151200.

9.

Find 9P3

A.) 414
B.) 514
C.) 504
D.) None of these

9P3 = 9!/6! = 9 × 8 × 7
= 504.

10.

Find 7P7

A.) 4440
B.) 5040
C.) 5045
D.) None of these

7P7 = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

11.

Find 4C2?

A.) 5
B.) 6
C.) 8
D.) 9

4C2
= 4!/(4-2)!2!
= 4!/(2! × 2!)
= (4 × 3)/2 = 6.

12.

Find 102C99?

A.) 71700
B.) 17100
C.) 17170
D.) 171700

102C99
= 102!/(102-99)!99!
= 102!/(3! × 99!)
= (102 × 101 × 100)/(3 × 2)

34 × 101 × 50 = 171700.

13.

In how many different number of ways the letters of the word 'ENGINEERING' can be arranged.

A.) 27200
B.) 27700
C.) 277200
D.) 27720

11!/(3! × 3! × 2! × 2! × 1!)
= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 3 × 2 × 2 × 2 × 1)
= 11 × 10 × 8 × 7 × 5 × 3
= 277200

14.

In how many different number of ways a combination of 3 persons can be selected from 4 men and 2 women.

A.) 150
B.) 200
C.) 250
D.) None of these

6C3 × 5C2
6C3
= 6!/(3! . 3!)
= (6 × 5 × 4)/(3 × 2)
= 5 × 4 = 20.
5C2
= 5!/(3! . 2!)
= 5 × 2 = 10
= 20 × 10 = 200.

15.

In how many different number of ways a combination of 3 men and 2 women can be selected from 6 men and 5 women.

A.) 15
B.) 20
C.) 25
D.) None of these

6C3
= 6!/(3! . 3!)
= (6 × 5 × 4)/(3 × 2)
= 5 × 4 = 20.

16.

In how many different number of ways a Committee of 3 person of can be selected from 5 men and 3 women such that atleast 1 women is included in the committee.

A.) 10
B.) 46
C.) 56
D.) None of these

1W 2M      2W 1M           3W
= 3C1 × 5C2 + 3C2 × 5C1 + 3C3s
= 3 × (5 × 4)/2 + 3 × 5 + 1
= 30 + 15 + 1 = 46
Total 5M 3W
8C3 = 56
5C3 = 10
Atleast one women = total - with out women
Atleast one women = 56 - 10 = 46

17.

In how many different ways can the letters of the word 'SCHOOL' be arranged so that the vowels always come together?

A.) 260
B.) 240
C.) 120
D.) None of these

'SCHOOL' contains 6 different letters.
vowels OO are always together
we have to arrange the letters (OO)SCHL
Now 5 letters can be arranged in 5! = 120 ways
The vowels (OO) can be arranged 2! = 2 ways.
= (120 x 2) = 240
again you have to divided 2 common vowels so answer is 120.