How many different committees of 2 members can be formed from a group of 7 people?

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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