# What is wrong? A committee of $6$ people is to be chosen from a group consisting of $7$ men and $8$ women.

A committee of $6$ people is to be chosen from a group consisting of $7$ men and $8$ women. If the committee must consist of at least $3$ women and at least $2$ men, how many different committees are possible?

The correct answer in the book is $3,430$, which they explain as $\dbinom{7}{3} \dbinom{8}{3} + \dbinom{7}{2} \dbinom{8}{4}$. Forming all of the committees with $3$ men and $3$ women, and then adding that to all of the committees with $4$ women and $2$ men.

My way of thinking about this is: $10 \cdot \dbinom{7}{2} \dbinom{8}{3}$. First pick $2$ men, then pick $3$ women, and then there are $10$ people left and one spot on the committee, so multiply all that by $10$. But this gets a different answer, $11,760$.

I understand how the book got its answer, but I'm wondering why my way of thinking about things is wrong.

• The probability tag probably doesn't need to be here. – G-man Mar 12 '15 at 17:14

## 3 Answers

Your answer is not correct because you are counting the same committee multiple times (once for each choice of who the "extra" person will be, among the gender with an extra). So you're counting each committee with $3$ men and $3$ women three times, and counting each committee with $2$ men and $4$ women four times. That's why you got $$3{{7}\choose{3}}{{8}\choose{3}} + 4{{7}\choose{2}}{{8}\choose{4}}=11760$$ instead of the correct answer of $${{7}\choose{3}}{{8}\choose{3}} + {{7}\choose{2}}{{8}\choose{4}}=3430.$$

• good job! +1.... – RE60K Mar 12 '15 at 17:34

You are doing, First :select something:, Second :Select something:. Since order is involved, you counted some committees thrice, some quadrice.

## Example

Select 2 men: $M_1,M_2$ Select 3 women: $W_1,W_2,W_3$ Select 1 another person: $M_3$ $$\text{is same as}$$ Select 2 men: $M_1,M_3$ Select 3 women: $W_1,W_2,W_3$ Select 1 another person: $M_2$

So you instead counted permutation of a subset of people as distinct comittees: $$\underbrace{\frac{3!}{2!}}_{\text{thrice}}\binom73\binom83+\underbrace{\frac{4!}{3!}}_{\text{quadrice}}\binom72\binom84$$ rather than: $$\binom73\binom83+\binom72\binom84$$

You've overcounted. For instance a committee of Alice, Betty, Carol, Dan, Ed, Frank could have been chosen as Alice, Betty, Carol Dan, Ed (as the initial five), and Frank (as the extra); or as Alice, Betty, Carol, Dan, Frank (as the initial five), and Ed (as the extra).

• No the last set is not possible, so this statement is wrong. – RE60K Mar 12 '15 at 17:33
• Right. Sorry. Should be Ed as the extra. I will fix it. – paw88789 Mar 12 '15 at 17:34
• ok! ${}{}{}{}{}{}$ – RE60K Mar 12 '15 at 17:35