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

Answer: Option ''

Answer: Option 'C'
npn = n! 
6p6
= 6 × 5 × 4 × 3 × 2 × 1 = 720

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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

Answer: Option 'D'

Answer: Option 'D' 
7p3 = 7 × 6 × 5 = 210

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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

Answer: Option 'A'

Answer: Option 'A'

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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

Answer: Option 'A'

Answer: Option 'A' 
The total number of arrangements is 
5P5  = 5! = 120

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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

Answer: Option 'C'

Answer: Option 'C' 
Total no'of arrangements 5P5  = 5! = 120 
no'of arrangements start with AE = 1 × 6 = 6 
no'of arrangements which do not start with AE = 120 - 6 = 114.

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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

Answer: Option ''

Answer: Option 'B' 
E,I fixed. Consonants can be arrangements in 3P3 = 3! = 6 ways

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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

Answer: Option 'A'

Answer: Option 'A'
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.

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8.

Find 10P6 

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

Answer: Option ''

Answer: Option 'B'
10P6 = 10!/4! = 10 × 9 × 8 × 7 × 6 × 5 
= 151200.

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9.

Find 9P3

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

Answer: Option 'B'

9P3 = 9!/6! = 9 × 8 × 7
= 504.

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10.

Find 7P7

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

Answer: Option 'B'

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

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11.

Find 4C2?

   A.) 5
   B.) 6
   C.) 8
   D.) 9

Answer: Option 'B'

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

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12.

Find 102C99?

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

Answer: Option 'B'

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

34 × 101 × 50 = 171700.

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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

Answer: Option 'C'

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

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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

Answer: Option 'B'

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

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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

Answer: Option 'B'

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

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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

Answer: Option 'B'

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

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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

Answer: Option 'C'

'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.

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Permutations Combinations Download Pdf

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