# Combinatorics- Why is this approach wrong?

In how many ways a committee of 6 members to be formed from 8 men and 5 women such that there are at least 2 men and 3 women?

Let men be $$m$$ and women be $$f$$.

I got the answer in second try using cases ($$4f+2m$$ or $$3f+3m$$) but in my first try I did something like this: $${8 \choose 2}{5 \choose 3}{8+5-5 \choose 1}$$.

According to me this should have given the right answer but it didn't.

Can someone please indicate what is the mistake I have made?

• What was the reasoning for your " first try" answer? – coffeemath Jan 22 at 12:23
• @coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1) – Sunil Kumar Jha Jan 22 at 12:25
• That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice. – coffeemath Jan 22 at 12:28
• @coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more? – Sunil Kumar Jha Jan 22 at 12:29
• I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen. – coffeemath Jan 22 at 12:32

By using the formula $$\binom{8}{2}\cdot \binom{5}{3}\cdot \binom{8+5-5}{1}$$ you are count the same committee more than one time. Let $$m_1,\dots, m_8$$ the male members and $$f_1,\dots, f_5$$ the female members. Then the committee $$m_1m_2f_1f_2f_3f_4$$ is counted one time when you first choose $$f_1f_2f_3$$ and then $$f_4$$ and a second time when you first choose $$f_2f_3f_4$$ and then $$f_1$$.

On the other hand, as you already noted, if we consider the two possible cases $$2m+4f$$ or $$3m+3f$$, we obtain the formula $$\binom{8}{2}\cdot \binom{5}{4}+\binom{8}{3}\cdot \binom{5}{3}$$ which counts every committee one and only one time.

• okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times? – Sunil Kumar Jha Jan 22 at 12:34
• @SunilKumarJha Yes, exactly. – Robert Z Jan 22 at 12:35

I suspect factor $$C(8,2)$$ stands for the number of ways $$2$$ men can be selected out of $$8$$ and similarly $$C(5,3)$$ for the number of ways $$3$$ women can be selected out of $$5$$.

Then you probabibly reason: $$8+5-5=8$$ persons are left and out of them $$1$$ must be chosen, leading to factor $$C(8,1)$$.

This however gives multiple counting.

Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $$8$$.

This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $$8$$.

• Thanks, i understood . – Sunil Kumar Jha Jan 22 at 12:37