This question already has an answer here:

I know what the binomial coefficient means in combinatorics, e.g. $\binom6{2}$ means "6 choose 2" i.e. how many different subsets of size 2 can there be, out of a set of 6 elements.

But what does it mean when there is a comma in the bottom of the coefficient, for example $\binom6{2,2}$?

I saw this usage in the first answer here, and I'd never seen it before: Problem 7, Ch1 from Blitzstein and Hwang, Intro to Probability


marked as duplicate by Community Aug 10 '17 at 1:15

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  • $\begingroup$ math.stackexchange.com/questions/2312648/… ?? $\endgroup$ – user451844 Aug 10 '17 at 1:07
  • $\begingroup$ en.wikipedia.org/wiki/Multinomial_theorem $\endgroup$ – Donald Splutterwit Aug 10 '17 at 1:07
  • $\begingroup$ @iamwhoiam see my link above found it using google search. $\endgroup$ – user451844 Aug 10 '17 at 1:09
  • $\begingroup$ Sorry guys, I am really tired and forgot to google. I will delete this. $\endgroup$ – Stephen Aug 10 '17 at 1:13
  • $\begingroup$ Apparently I can't do that since it already has an answer. But feel free to do it yourselves. $\endgroup$ – Stephen Aug 10 '17 at 1:14

This is known as the multinomial coefficient:

$${k\choose t_1, t_2, ... , t_n }=\frac{k!}{t_1! \ldots t_n!}$$

As mentioned in the comments, we must have that $t_1+t_2+\dots +t_n = k$

  • $\begingroup$ It should be specified that this notation requires $t_1+t_2+\dots+t_n=k$. The notation $\binom{6}{2,2}$ is meaningless as far as I'm aware since $2+2\neq 6$. Also, it isn't the multinomial theorem, it is merely a multinomial coefficient. $\endgroup$ – JMoravitz Aug 10 '17 at 1:11
  • $\begingroup$ This is true. I will add to the answer $\endgroup$ – K Split X Aug 10 '17 at 1:13
  • $\begingroup$ If $\binom6{2,2}$ is meaningless, then that would imply the accepted answer at the question I linked is wrong. But it seems more like they mean something like "choose 2 from 6, then choose 2 again [from the remaining 4]", at least reading between the lines of the answer... $\endgroup$ – Stephen Aug 10 '17 at 1:30
  • $\begingroup$ If you want to be pedantic, we can say $\binom{6}{2,2}$ is equal to zero since there are no ways to partition a set of $6$ elements into an ordered partition of partsizes $2$ and $2$. $\endgroup$ – JMoravitz Aug 10 '17 at 1:38

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