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So there are two committees, committee $A$ and committee $B$. committee $A$ will have $5$ members, committee $B$ will have $6$. people can be on both committees. There are $n \geq 6$ people to choose from. What is the total number of ways the committees can be chosen?

So if $n$ was a constant number, say $6$, I would say that for committee $A$: $$\binom{6}{5}$$ and for committee B: $$\binom{6}{6}$$ and then multiply the two answers.

if $n=7$, similarly committee $A$: $$\binom{7}{5}$$ and for committee B: $$\binom{7}{6}$$

But the question asks me:

What is the total number of ways in which these two committees can be chosen when $n \geq 6$ ? Justify your answer

Would putting $$\binom{n}{5} * \binom{n}{6}$$ Be too vague of an answer ? What am I missing here?

Olórin
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saeid
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  • $n$ will have to be in your answer, so this would be correct. Maybe write it out in terms of factorials and fractions to see if there is any canceling out available. – turkeyhundt Jan 30 '15 at 01:26
  • The answer is correct. Technically, the answer is also correct for $n\lt 6$, because it is common to define $\binom{n}{6}$ to be $0$ in that case. But since obviously no way to do the job if $n\lt 6$, the question specified $n\ge 6$. – André Nicolas Jan 30 '15 at 01:29

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