Q:

Permuting men and women in Supreme Court

I ask this question mainly to resolve (hopefully) and error with the following problem.  
The United States Court consists of $3$ women and $6$ men.  In how many ways can a $3$-member committee be formed if each committee must have at least one woman? 
My approach:
Since each group needs at least one woman, there are $\binom31$ ways to pick the first one.  After that, both men and women are allowed.  There are $\binom82$ ways of doing this.  
This amounts to $\binom31\times\binom82 = 84$ ways, but the answer is $64$. I don't see how this is the answer for two reasons.  The first is that there should be at least somewhere a $3$ multiplied but $64$ has no powers of $3$. The second is that I wrote a counter in Java to look at all strings from the list $a,b,c,d,e,f,g,h,$ and $i$ choosen $3$ at a time and count how many words contained $a,b,$ or $c$. 
What seems to be the problem? 

A:

Your problem double counts situations where say, Elena was your $3\choose1$ choice and Ruth was in the $8\choose2$ field (plus, let's say, Sam) and then vice versa. 
A better choice when you see the phrase "at least" is to find the opposite situation and subtract from the full options.   In this case any committee of 3 minus a committee of 3 men. 
$${9\choose3}-{6\choose3}=64$$

A:

because you are overcounting.   Suppose you select woman 'A' first, other 2 two woman 'B' and 'C' may appear in the next choice.  Suppose you select woman 'B' first, other 2 two woman 'A' and 'C' may appear in the next choice.  likewise, over-counting happens.  
Case1: exactly one woman
$\dbinom{3}{1}\times \dbinom{6}{2}=45$
Case2: exactly two woman
$\dbinom{3}{2}\times \dbinom{6}{1}=18$
Case1: exactly three woman'
$\dbinom{3}{3} =1$
So, total is $64$
But turkeyhundt's approach is better

