There are $10$ men and $7$ women working as supervisors in a company. The company has recently decided to form a committee to represent all the employees. The committee has to consist of $3$ members, all of whom must be supervisors. The members will be President, General Secretary and Coordinator respectively. Answer the following questions based on this information.

(a) How many ways can the committee be formed from the supervisors available?
(b) How many ways can the committee be formed if the General Secretary must be a female?
(c) How many ways can the committee be formed if it must have at least one man and at least one woman?

(a) $\binom{17}{3}$
(b) $\binom{16}{2}$
(c) $\binom{10}{1}\binom{7}{2} + \binom{10}{2} \binom{7}{1}$.

Do you think my answers are correct?

  • $\begingroup$ Please format your questions into logical partitions. $\endgroup$
    – evil999man
    Apr 21, 2014 at 18:00

2 Answers 2


For $(a)$ you have $17\cdot 16\cdot 15=4080$ options.

For $(b)$ you have $7$ options for the General Secretary, times $16\cdot 15=240$, which gives $1680$.

For $(c)$ just multiply your answer by $3!$ to account for the fact that the positions are labeled.


Your answers look mostly correct. There is an easier way to do $(c)$ though. Start by picking a man and a woman. That gives us $\binom{10}{1} \cdot\binom{7}{1} = 70$. We then choose one person out of the $15$ remaining. We then have to permute the members, so we multiply by $3!$. So our answer is $70 \cdot15 \cdot3!$. However, we also have to divide out by $2$ to handle the symmetry cases. So if $m_{1}$ is drawn first and $m_{2}$ is drawn second, the permutations handle the ordering of those.

Note- for all of your answers, since the committees have labeled positions, you have to permute the members through the positions. So you have to multiply by $3!$ for each of your solutions.

  • $\begingroup$ Well, I see how your approach to (c) is a whole lot easier. But, it turns out it returns an answer that's double of mine. Not sure where I went wrong. $\endgroup$ Apr 21, 2014 at 19:20
  • $\begingroup$ Did you revise your answer to permute the elements? Your answer for (c) isn't handling ordering. $\endgroup$
    – ml0105
    Apr 21, 2014 at 19:26
  • $\begingroup$ I understand that I do have to permute the elements, but assuming for the moment that the question has three equal committee members, my answer and yours still don't match. I have $\binom{10}{1}\binom{7}{2} + \binom{10}{2} \binom{7}{1}$ whereas you have (70)(15). That's where I'm stuck. $\endgroup$ Apr 21, 2014 at 19:34
  • $\begingroup$ I forgot to divide out by $2$ so I don't count permutations of members of a single group twice. I've updated my answer, and I apologize for any confusion I may have caused. $\endgroup$
    – ml0105
    Apr 21, 2014 at 20:04

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