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How many different combinations of $3$ can you make with $11$ items?

I would think the answer to be $11\cdot10\cdot9$ but this is incorrect.

Thanks.

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    You have counted every combination $6$ times, as $A$ then $B$ then $C$, also as $A$ then $C$ then $B$, also $\dots$.2012-12-11
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    Ah, thanks, what is the general method to doing these types of questions?2012-12-11
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    Look up "binomial coefficients" or "[combinations](http://en.wikipedia.org/wiki/Combination)".2012-12-11
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    Good answer given by Brian M. Scott.2012-12-11

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Suppose that the items are the first $11$ letters of the alphabet. The set $\{A,B,C\}$ is one combination, but you’ve counted it six times, once for each of the $3!=6$ possible orders in which you could have picked it ($ABC,ACB,BAC,BCA,CAB$, and $CBA$). To get the number of combinations (as distinct from permutations), you must divide by $6$.

More generally, the number of $k$-element combinations from a set of $n$ objects is $$\binom{n}k=\frac{n!}{k!(n-k)!}\;,$$ not $$n(n-1)\dots(n-k+1)=\frac{n!}{(n-k)!}\;,$$ which is what you computed. That extra factor of $k!$ in the denominator is the number of different orders in which you can pick a set of $k$ things; dividing by it gives you the number of actual combinations (subsets), rather than the number of different lists (permutations) of $k$ objects.

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    Thanks alot! Excatly what i was looking for2012-12-11
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    SO the answer is 165?2012-12-11
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    @fosho: Yes, it’s $165$. And you’re welcome!2012-12-11