There are $5$ women, $3$ men. How many ways to form a committee of $3$ with at least $1$ member of the opposite sex?

I have looked through several topics for similar solutions and I have attempted an answer to the question. Unfortunately, the sample question itself does not have an answer.

From $5$ women and $3$ men, how many ways are there to form a committee of $3$ where there has to be at least $1$ member of the opposite sex.

My first attempt:

$$\binom{5}{3} \cdot \binom{3}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = 10 \cdot \frac{3!}{3!}$$

Since there are only two ways for committees with no same gender, I did $10-2 = 8$ as my final answer.

My second attempt:

First case: There has to be at least $1$ male. So that means that there are $\binom{5}{2}$ women available. $\binom{5}{2}$ is $10$. $\binom{3}{1}$ is $3$, so $10 \cdot 3 = 30$.

Second case: There has to be at least $2$ males. So that means $\binom{5}{1}$ women are available. $\binom{5}{1}$ is $5$, and $\binom{3}{2}$ is $3$. So that means there are $15 \cdot 3 = 45$ choices available.

When you add the two choices, you get $30 + 45 = 75$ choices.

Which attempt is right? Or are they both wrong? I would love explanations because the sample question has no answer listed.

• I think you clarified it in the discussion, but just to be sure: the condition is just that you exclude $3$ women or $3$ men, yes? – lulu Mar 21 '16 at 21:28
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 21 '16 at 23:10
• You made a typographical error in your second case in your second attempt, which led to the wrong answer. – N. F. Taussig Mar 21 '16 at 23:12

So $$\{\mbox{committees with at least one member of hte opposite sex}\} = \{\mbox{all possible committees}\} - \{\mbox{committees of all the same sex}\}$$ As "all the same sex" could be either "all men" or "all women," adding it up, we get a total of $$\binom{8}{3} - \bigg(\binom{5}{3} + \binom{3}{3}\bigg).$$
There are only two possibilities: (i) 2 women and 1 man and (ii) 1 woman and 2 men. This is simply $$\binom{5}{2}\binom{3}{1}+\binom{5}{1}\binom{3}{2}=45.$$ Neal's hint gives the same solution, and is computationally simpler (hence superior). You might have an easier time understanding the above, however.
Your first method is conceptually sound, but computationally flawed. With no restrictions on gender we just need to select $3$ people from the available $8$, hence $\binom 83=56$ possible choices. Now, exactly one of these is all male but there are $\binom 53=10$ ways to assemble an all female group. Hence $$56-10-1=\fbox {45}$$
Your second method is also flawed, though again your thinking is fine. To make a committee with $2$ women and $1$ man we have $\binom 52\times \binom 31 = 10\times 3=30$ possibilities. To make a committee with $1$ woman and $2$ men we have $\binom 51\times \binom 32 = 5\times 3=15$ possibilities. And again we get $$30+15=\fbox {45}$$
One more formulation. From the number of all possible committees, substract the number of committees with only men and the number of committees with only women. That is : $$\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45$$