0
$\begingroup$

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.

  • 0
    Good answer given by Brian M. Scott.2012-12-11

1 Answers 1

0

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.

  • 0
    @fosho: Yes, it’s $165$. And you’re welcome!2012-12-11