Permutation and Combination
Permutation and Combination - Important Points
| 26. | Out of 2 men, 3 women, 4 girls and 5 boys, in how many ways can we form a group of three people such that a boy and a girl will not be together in the group? |
|---|
A. 210
B. 204
C. 168
D. 194
View Answer Discuss Work SpaceAnswer: option d
Explanation:
Since, boys and girls are not taken together, so there are 2 possible ways, i.e., to choose from (2 men, 3 women and 4 girls) or (2 men, 3 women and 5 boys).
So, the number of possible ways = 9C3 + 10C3
But, all the 3 member formed by 2 men and 3 women is counted twice. So we need to subtract 5C3 from the total number of ways.
Therefore, the answer would be 204 – 10 = 194
| 27. | Find the number of squares on a chessboard? |
|---|
A. 204
B. 200
C. 256
D. 128
View Answer Discuss Work SpaceAnswer: option a
Explanation:
There will be 64 squares of size 1 × 1, 49 squares of size 2 × 2, 36 squares of size 3 × 3 and so on till 1 square of size 8 × 8.
Hence, the total number of squares
= 64 + 49 + 36 + … + 1 = 204.
Alternatively,
Number of squares in a square grid of size n x n = ∑n2 = n (n+1) (2n+1) / 6
now putting n = 8 we get 8 x 9 x 17 / 6 = 204
| 28. | In how many ways, 6 girls may be selected for a team out of 12 girls in a class so that 2 particular girls (captain and vice-captain) are always there in the team? |
|---|
A. 360
B. 210
C. 240
D. 120
View Answer Discuss Work SpaceAnswer: option b
Explanation:
Total number of girls = 12
Number of girls to be selected = 6
If the captain and vice-captain are already selected, then it is required that 4 girls out of 10 have to be selected.
So, Number of ways = 10C4 = 210
| 29. | In an entrance examination paper, 5 questions on Maths were asked. In how many ways students could attempt paper such that at least one question has to be attempted? |
|---|
A. 120
B. 63
C. 32
D. 31
View Answer Discuss Work SpaceAnswer: option d
Explanation:
Number of Maths questions in the paper = 5
An attempt of at least one question = one question or two or three or four or five questions attempt.
=5C1 + 5C2 + 5C3 + 5C4 + 5C5
= 5 + 10 + 10 + 5 + 1 = 31
Alternatively,
Number of ways = (2)^5 – 1 = 32 – 1 = 31
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