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Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 1)
Permutation and Combination
1.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
564
645
735
756
None of these
Answer: Option
Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2)
3 x 2 x 1 2 x 1
= 525 + 7 x 6 x 5 x 6 + 7 x 6
3 x 2 x 1 2 x 1
= (525 + 210 + 21)
= 756.

Discussion:
137 comments Page 14 of 14.
Sumanth reddy said:   2 years ago
How it's changed from 7c4 to 7c3? Please explain me.
(23)
Vishwanath said:   1 year ago
We need a minimum 3 men, so the maximum can be all 5 men in the committee.

3 men AND 2 women = 7C3 X 6C2 = 525.
4 men AND 1 women =7C4 X 6C1 = 210.
all 5 men = 7C5 = 21.

Add them 525 + 210 + 21 = 756.
(25)
Khan said:   1 year ago
Why can't it be, 7C3 (at least 3 men) * 10C2 (any 2 may be men or women or both out of the remaining 10)?
(21)
Adith said:   1 year ago
Thanks everyone for explaining the answer.
(9)
Abhishek Singh said:   7 months ago
3 men & 2 woman = 7C3 * 5C2 = 525.
4 men & 1 woman = 7C4 * 5C1 = 210.
5 men = 7C5 = 21.
So, the answer is 756.
(6)
Kavya said:   7 months ago
How to solve this? Please explain.
(5)
Melvin N said:   3 months ago
nCr = nC(n-r) --> formula.

7C3 = 7C(7-3) = 7C4 = 35.
(2)
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