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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 3 of 14.
Anand kumar Gupta said:   6 years ago
Here 5 persons are select, and at least 3 men including this committee so we have
Men. Women
7c3 * 6c2 = 525
7c4 * 6c1 = 210
7c5 * 6c0 = 021
= 756 is the answer.
(1)
Johar said:   4 years ago
Anyone, Please solve this problem clearly to get it.
(1)
Melvin N said:   1 month ago
nCr = nC(n-r) --> formula.

7C3 = 7C(7-3) = 7C4 = 35.
(1)
Naveen said:   2 decades ago
First of all we need to select 3 man from 7 so = 7C3 = 35
Then we can choose 2 person from (7-3)+ 6 person ie. 10C2 = 45
So answer should be 53*45 = none of these..

Please tell me where I'm wrong.
Kumar said:   1 decade ago
Nice answer nagu.
Nikhil said:   1 decade ago
Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2. Please Clarify.
Madhusudan said:   1 decade ago
In Above solution I am unable to understand why have we included 5/1(in both the committee and 6/2 in 2nd committee. Can any help me to understand how the combination considered in the example.
Anu said:   1 decade ago
5 ball are to be placed in 12 boxes, they are placed in three rows such that each row contain at least one ball in how many ways it can be placed.
Bhargav said:   1 decade ago
Not a good answer to give please another method please easy method.
Bhargav said:   1 decade ago
How can you take a 3 men 2 women 4men and 1women and a 5 men please solve easy method.
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