<s>[INST] 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.
 [/INST] 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 :
$$
\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45
$$
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