Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 11)
11.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Answer: Option
Explanation:
| Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = |  |
7 x 6 | x 3 |  |
= 63. |
| 2 x 1 |
Discussion:
51 comments Page 5 of 6.
Srikanth said:
1 decade ago
@teja. As this a problem f selection you use combination. It means calculating 7c5 is a bit more tedious when compared to calculating 7c2.
Moni said:
1 decade ago
For ex. Take 7C5, 5 is more than a half of 7. So we can use the formula nC (n-r) when are is more than half of n.
Similarly 3C2 is made as 3C1. I hope you have understood.
Similarly 3C2 is made as 3C1. I hope you have understood.
Niyati Bafna said:
1 decade ago
Are there any more problems, perhaps of a slightly advanced kind, on the net? If so, on which sites can they be found?
Vinod Rathod said:
1 decade ago
Lets do by this way :
As you said that (7c5*3c2) = (7c2*3c1).
7/5*3/2 = 7/2*3/1 = 63.
As you said that (7c5*3c2) = (7c2*3c1).
7/5*3/2 = 7/2*3/1 = 63.
Pooja said:
1 decade ago
I understood that thanks! we have to use nCr=nCn-r.
Chandan said:
1 decade ago
Why multiplication not addition?
Aswin said:
1 decade ago
Another question : how to select 6 people from a group of 10 so that two of them does not come together?
Shital said:
1 decade ago
I didn't understand 7c2*3c1.
Vishesh said:
1 decade ago
Same here what is this c? I don't understand this.
Manju said:
1 decade ago
nCr = n!/(n-r)!r!
= 7C5*3C2.
= (7*6/2*1)*3 = 63.
= 7C5*3C2.
= (7*6/2*1)*3 = 63.
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