Let $1,2,3,4,5,6,7,8$ be the women, and the men are denoted similarly, but with a hat, $\hat1$, $\hat2$, $\hat3$, $\hat4$, $\hat5$, $\hat6$.
The $1$ is not willing to work with $\hat 1$.
Now let us look at the counting strategy. Where do we count the configuration $2,3,4; \hat2,\hat3,\hat 4$?
- Well, at point one at any rate, it is one valid case among the many $\binom 73\binom 53$, since $1$ and $\hat 1$ are both excluded.
- Well, at point two also, it is one valid case among the many $\binom 83\binom 52$ cases, since $\hat 1$ is excluded.
- Well, at point three again, it is one valid case among the many $\binom 72\binom 63$ cases, since $1$ is excluded.
For a correct answer, count all possible commitees, there are $\binom 83\binom 63=1120$ of them, and subtract those where the pair $1,\hat 1$ is in the commitee, there are $\binom 72\binom 52=210$ of them. So the answer is $$ \binom 83\binom 63 - \binom 72\binom 52 =1120 -210 =910\ . $$
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How many different committees of 2 men and 2 women can be formed from [#permalink]
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How many different committees of 2 men and 2 women can be formed from a group of 12 people, half of whom are men?
A. 225
B. 450
C. 495
D. 900
E. 2,970
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Re: How many different committees of 2 men and 2 women can be formed from
[#permalink]
Bunuel wrote:
How many different committees of 2 men and 2 women can be formed from a group of 12 people, half of whom are men?
A. 225
B. 450
C. 495
D. 900
E. 2,970
6 men and 6 women to choose from
2 men and 2 women to be chosen
Total Selections = 6C2*6C2 = 15*15 = 225
Answer: Option A
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Re: How many different committees of 2 men and 2 women can be formed from [#permalink]
6 Men 6 Women
Total no of ways
6C2 * 6C2
= [6*5/2] * [6*5/2]
= 15*15
=225
IMO A
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Re: How many different committees of 2 men and 2 women can be formed from [#permalink]
Total number of people = 12
Men = 6 and Women = 6
No. of ways to form a committee with 2 Men and 2 Women = \[6C2 * 6C2\] = 15*15 = 225
Hence it's A
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Re: How many different committees of 2 men and 2 women can be formed from [#permalink]
Total no of ways to choose committee of 2 men and 2 women = 6C2*6C2 = 15*15 = 225
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Re: How many different committees of 2 men and 2 women can be formed from [#permalink]
Total: 12 => men : 6 and women: 6
Committtee: 2 men and 2 women: \[^6[C_2] * ^6[C_2]\]
=> 15 * 15 = 225
Answer A
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Re: How many different committees of 2 men and 2 women can be formed from [#permalink]
09 Oct 2020, 02:39
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