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?

  • 3
    $\begingroup$ No it isn't. It is 3:5 $\endgroup$ – Doug M Dec 20 '16 at 21:27
  • $\begingroup$ Did the Java counter also say 84? If so, you might want to try to find the error in it. $\endgroup$ – David K Dec 20 '16 at 21:45
  • $\begingroup$ The answers provided by turkeyhundt and kiran are correct. While you are counting each committee with one woman once, you are counting each committee with two women twice and each committee with three women three times, once for each way you could designate one of the women on the committee as the designated woman on the committee. Notice that $$\binom{3}{1}\binom{6}{2} + \binom{2}{1}\binom{3}{2}\binom{6}{1} + \binom{3}{1}\binom{3}{3}\binom{6}{0} = 84$$ $\endgroup$ – N. F. Taussig Dec 21 '16 at 1:44
  • $\begingroup$ How does one notice when there is an overcount? How do you prevent this from happening? $\endgroup$ – Ian Limarta Dec 21 '16 at 5:07

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.



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


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