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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 10 of 14.
Rajeev said:   1 decade ago
In the first step 7c5 converted 7c2 in the second step. But how ?
Miguel said:   1 decade ago
If nCr=nCn-r, then why isn't 7c3 (from 7c3 x 6c2) changed to 7c4?
Vibhor said:   1 decade ago
How to know when to use permutation and when to use combination ?
Ashok said:   1 decade ago
Please give a correct explanation to combination and permutation?
Divya said:   1 decade ago
May be 5 womans is also possible, why we don't take that option?
Maryann said:   1 decade ago
Can someone explain better for me because I'm getting confused?
Vinay teja said:   6 years ago
There are 5members required so why don't we divide with 11c5?
Ishika Singh said:   7 years ago
I am not getting the solution. Please, anyone explain to me.
Santosh said:   1 decade ago
It is so long procedure, is any shortcut method for this?
D.Yogapriya said:   1 decade ago
Is there any other simple way to solve this 1st problem?
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