Aptitude - Permutation and Combination
Overview Exercise "Example is better than precept."
- (Proverb)
6.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Answer: A
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
= (6 C1 x 4 C3 ) + (6 C2 x 4 C2 ) + (6 C3 x 4 C1 ) + (6 C4 )
= (6 C1 x 4 C1 ) + (6 C2 x 4 C2 ) + (6 C3 x 4 C1 ) + (6 C2 )
= (6 x 4) +
6 x 5
x
4 x 3
+
6 x 5 x 4
x 4
+
6 x 5
2 x 1
2 x 1
3 x 2 x 1
2 x 1
= (24 + 90 + 80 + 15)
= 209.
7.
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer: D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
8.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Answer: B
Explanation:
Required number of ways
= (8 C5 x 10 C6 )
= (8 C3 x 10 C4 )
=
8 x 7 x 6
x
10 x 9 x 8 x 7
3 x 2 x 1
4 x 3 x 2 x 1
= 11760.
9.
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Answer: B
Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
= (3 C1 x 6 C2 ) + (3 C2 x 6 C1 ) + (3 C3 )
=
3 x
6 x 5
+
3 x 2
x 6
+ 1
2 x 1
2 x 1
= (45 + 18 + 1)
= 64.
10.
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
Answer: E
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3 P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3 P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
Required number
