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.
 A: Hint: The opposite of "at least one member of the opposite sex" is "all members of the same sex."

 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 $$\binom{8}{3} - \bigg(\binom{5}{3} + \binom{3}{3}\bigg).$$

A: 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.
A: 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 $$30+15=\fbox {45}$$
A: 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 :
$$
\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45
$$
