Let $X$ be a set of $n$ elements. Suppose that we have two colors of paint, say red and blue, and we are going to choose a subset of $k$ elements to be painted red with the rest painted blue. Then the number of different ways this can be done is just the binomial coefficient $\binom{n}{k}\text{.}$ Now suppose that we have three different colors, say red, blue, and green. We will choose $k_1$ to be colored red, $k_2$ to be colored blue, and the remaining $k_3 = n - (k_1+k_2)$ are to be colored green. We may compute the number of ways to do this by first choosing $k_1$ of the $n$ elements to paint red, then from the remaining $n-k_1$ elements choosing $k_2$ to paint blue, and then painting the remaining $k_3$ elements green. It is easy to see that the number of ways to do this is \begin{equation*} \binom{n}{k_1}\binom{n-k_1}{k_2} = \frac{n!}{k_1!(n-k_1)!} \frac{(n-k_1)!}{k_2!(n-(k_1+k_2))!} = \frac{n!}{k_1!k_2!k_3!} \end{equation*} Numbers of this form are called multinomial coefficients; they are an obvious generalization of the binomial coefficients. The general notation is: \begin{equation*} \binom{n}{k_1,k_2,k_3,\dots,k_r}=\frac{n!}{k_1!k_2!k_3!\dots k_r!}. \end{equation*}

For example, \begin{equation*} \binom{8}{3,2,1,2}=\frac{8!}{3!2!1!2!}= \frac{40320}{6\cdot2\cdot1\cdot2}=1680. \end{equation*}

Note that there is some “overkill” in this notation, since the value of $k_r$ is determined by $n$ and the values for $k_1\text{,}$ $k_2,\dots,k_{r-1}\text{.}$ For example, with the ordinary binomial coefficients, we just write $\binom{8}{3}$ and not $\binom{8}{3,5}\text{.}$

##### Example2.32

How many different rearrangements of the string: \begin{equation*} \text{MITCHELTKELLERANDWILLIAMTTROTTERAREREGENIUSES!!} \end{equation*} are possible if all letters and characters must be used?

Solution

Just as with binomial coefficients and the Binomial Theorem, the multinomial coefficients arise in the expansion of powers of a multinomial:

##### Example2.34

What is the coefficient of $x^{99}y^{60}z^{14}$ in $(2x^3+y-z^2)^{100}\text{?}$ What about $x^{99}y^{61}z^{13}\text{?}$

Solution