There are $5$ senior students and $3$ junior students. They are to form a committee of $5$, in which $3$ are decision makers and $2$ handle logistics. Only seniors can be decision makers. But anyone can handle the logistics.

How many possible combinations (order does not matter within the two designations) are there?


Out of $5$ seniors, we choose $3$ to be decision makers. Then, in the remaining $5$ students, we choose $2$ to handle logistics. Thus, the answer is: $5 \choose 3$$5 \choose 2$.


Is my reasoning correct? (is this kind of question allowed here?)

  • 1
    $\begingroup$ Yes, the answer should be correct $\endgroup$ – malin Nov 3 '12 at 13:34
  • $\begingroup$ (and this kind of question is OK.) $\endgroup$ – M.B. Nov 3 '12 at 13:34
  • $\begingroup$ Thanks. Good to know its allowed. Didn't want to ask without posting an attempt. I suppose I'll answer my question with "yes my reasoning is correct"? $\endgroup$ – Legendre Nov 3 '12 at 13:38
  • $\begingroup$ The thing to remember is the more constraint you have on a choice to make, the more you want to make that choice first $\endgroup$ – Jean-Sébastien Nov 3 '12 at 14:30
  • $\begingroup$ There is one point not clear in the question: does the choice only involve selecting the committee (which must have at least three seniors), or must it also involve designating roles within the committee? In some possible committees there are more than three senior students; for those do we already have to tell who will be decision makers? Only if the latter is required is your answer correct. $\endgroup$ – Marc van Leeuwen Nov 3 '12 at 14:46

Based on the comments, my reasoning is correct.

  • $\begingroup$ Your answer is incorrect. We must select how many seniors are on the committee, which seniors occupy decision-making positions, and which juniors are on the committee. $$\binom{5}{3}\binom{3}{3}\binom{2}{2} + \binom{5}{4}\binom{4}{3}\binom{2}{1} + \binom{5}{5}\binom{5}{3}\binom{2}{0}$$ $\endgroup$ – N. F. Taussig Apr 10 '16 at 12:19

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