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In an answer to a question (https://math.stackexchange.com/a/288999/172737) this combinatoric symbol was used

$$\binom {i+j+k}{i,j,k}$$

I can't determine its meaning, though I've searched.

I guessed it's meant to be a product of the combinations using the denominators in turn in the usual symbol, but assuming this does not give me the answer to case $n=3$ in the answer of the link.

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  • $\begingroup$ It is $\dfrac{(i+j+k)!}{i!j!k!}$ $\endgroup$
    – ElementX
    Commented Nov 11, 2018 at 13:58

4 Answers 4

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This is a multinomial coefficient.

One defines, for $n= m_1+m_2 +m_3$: $$\binom{n}{m_1,m_2,m_3}= \frac{n!}{m_1!m_2!m_3!}$$ and more generally, for $n= m_1+m_2 +\dots +m_k$: $$\binom{n}{m_1,m_2,\dots,m_k}= \frac{n!}{m_1!m_2!\dots m_k!}$$

The binomial coefficient would be the special case $k=2$; to simplify notation one usually writes $\binom{n}{m}$ instead of $\binom{n}{m,n-m}$.

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Don't really worry too much about what it is. Worry about what it means:

Think about if I have 20 people, and I want to divide it in 8, 5,7, here is how I would do it:

$$\binom{20}{8}\binom{12}{5}\binom{7}{7}$$

But, this is frequently written as follows:

$$\binom{20}{\text{8 5 7}}=\dfrac{20!}{8!5!7!}$$

You can calcuate it in this way instead of doing choose.

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It is a multinomial coefficient: $$\binom {i+j+k}{i,j,k}=\frac{(i+j+k)!}{i!\:j!\:k!}$$ used in the multinomial formula for a sum of three terms: $$ (x+y+z)^n=\sum_{\substack{i,j,k\\i+j+k=n}}\binom{n}{i,j,k} x^i y^j z^k.$$

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It is the multinomial coefficient. Formula is factorial of upper sum, divided by product of the factorials of the numbers between commas in the "denominator". The usual binomial coefficient $\binom{n}{k}$ can then be also rewritten $\frac{n!}{k!\cdot (n-k)!},$

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