What does this combination notation mean?

Question

I feel stupid even asking this but when they have a combination like

$15\choose5,7,3$

does that just mean

$15\choose5$ $10\choose7$ $3\choose3$

Answer 1

Yes, it is actually what you think it is, they are called multinomial coefficients, but note what you wrote below is also equvalent to $\dfrac{15}{5!\cdot7!\cdot3!}$ in general $\binom{n}{m_1,m_2\dots m_j}=\dfrac{n!}{m_1!\cdot m_2!\dots\cdot m_j!}$. And this is of course the number of ways to color $n$ objects such that there are $m_1$ of the first color,$m_2$ of the second $\dots$

Answer 2

It may depend on context but the only time I've seen notation like that is when referring to the multinomial coefficient, which is defined by:

$$\binom{n}{k_{1},\dots,k_{n}}=\frac{n!}{k_{1}!\cdots k_{n}!}$$

It is interesting to note that we can write this as a product of binomial coefficients:

$$\binom{n}{k_{1},\dots,k_{n}}=\binom{k_{1}+k_{2}+\dots+k_{n}}{k_{2}+\cdots+k_{n}}\cdots\binom{k_{n-1}+k_{n}}{k_{n-1}}$$

This is used in the multinomial theorem which states:

$$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}$$

Answer 3

$$\require{cancel} \binom{15}{5}\binom{10}{7}\,\binom{3}{3} = \dfrac{15!}{5! \cancel{10!}}\cdot \frac{\cancel{10!}}{7!3!} \cdot 1 = \dfrac {15!}{5!7!3!} = \binom{15}{5, 7, 9}$$

So yes, what you post indeed holds, but the first expression, $\dbinom{15}{5, 7, 3}$ is a more concise way of stating what you wrote.