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.

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 :
$$
\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45
$$
