# 20 options. 10 must be chosen. how many combinations exist

I have a list of twenty options and each option is different. If ten options must be chosen, how many combinations exist? Can someone show me how the math on this?

Example, if i have 20 different colors of cards. How many combinations can I get from choosing ten of them. I hope that makes sense.

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Assuming that order doesn't matter, the answer is $$\binom{20}{10} = \frac{20!}{10! 10!}$$ where the left side is read as "20 choose 10."

The reasoning is as follows: We can choose 10 objects out of the 20 by putting all 20 objects in a random order and then taking the first 10 of them (with cards, for example, shuffle them together, then deal off the top 10). This gives us $20!$ possible orders. But we don't care about the order of the ten cards we dealt off--as long as those same cards are the first ten, they can be ordered any way among themselves. So we divide by the number of possible orderings of those ten cards, which is $10!$. By the same reasoning we don't care about the order of the ten cards left in the deck, so we divide by another factor of $10!$ for those.

In general, if you have $n$ objects and you want to choose $k$ of them, the number of possible ways to do that is $$\binom{n}{k} = \frac{n!}{k!(n-k)!}.$$

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Perfect, Thank you. – user903497 Oct 28 '12 at 22:34
This is assuming that the order of the ten does not matter right? Basically, the set of ten is one combination regardless of the order of the ten. With that said, the formula above still holds right? – user903497 Oct 28 '12 at 23:26

There are 184,756 combinations 10 things, given a set of 20 to choose from.

let me find the Tex for you. Basically, you divide 20! (factorial) by 10!^2

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can you show me how you calculated that? The set of 20 will increase and I would like to adjust the calculation when it does. Thank you! – user903497 Oct 28 '12 at 22:31
Thank you very much. Confirmed it on wolframm. +1 for speed of answer – user903497 Oct 28 '12 at 22:34
Well, all the big kids already got their Tex in. I guess it's up to you if you prefer speed over prettiness. – New Alexandria Oct 28 '12 at 22:38

Imagine that you choose your $10$ options sequentially. There are $20$ possibilities for the first option. After you’ve chosen it, $19$ possibilities remain for your second option. Continuing to reason in this fashion, we see that there are

$$20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11$$

ways to choose $10$ options in sequence; this number can be more compactly written $\dfrac{20!}{10!}$.

However, we’ve counted each possible set of $10$ options many, many times: we’ve counted it once for every possible order in which we could have picked it. How many times is that?

Suppose that we have a set of $10$ things, and we line them up. We can choose any of the $10$ to go first; once that’s been done, we can choose any of the $9$ remaining objects to go second; and so on. Thus, we can arrange the $10$ objects in any of $$10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=10!$$ orders.

Going back to the original problem, this means that the figure of $\frac{20!}{10!}$ counts each $10$-option set $10!$ times, once for each of orders in which we could have selected it, i.e., once for each of its possible permutations. The number of different sets of $10$ options is therefore only $\frac1{10!}$-th of $\frac{20!}{10!}$, or $\frac{20!}{10!10!}$. This number is the binomial coefficient

$$\binom{20}{10}=\frac{20!}{10!10!}\;.$$

The same reasoning applied to the general problem of counting the $k$-element subsets of a collection of $n$ things leads to the binomial coefficient

$$\binom{n}k=\frac{n!}{k!(n-k)!}\;.$$

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