Basic Computer Knowledge Test Questions and Answers

Permutations Combinations

1.

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'

2.

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

3.

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.

4.

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

5.

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.

6.

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.

7.

Find 9P3

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

Answer: Option 'B'

9P3 = 9!/6! = 9 × 8 × 7
= 504.

8.

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

9.

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.

10.

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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