## Section2.7Multinomial Coefficients

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{MITCHELTKELLERANDWILLIAMTTROTTERAREGENIUSES!!} \end{equation*}

are possible if all letters and characters must be used?

Solution

To answer this question, we note that there are a total of $45$ characters distributed as follows: 3 A's, 1 C, 1 D, 7 E's, 1 G, 1 H, 4 I's, 1 K, 5 L's, 2 M's, 2 N's, 1 O, 4 R's, 2 S's, 6 T's, 1 U, 1 W, and 2 !'s. So the number of rearrangements is

\begin{equation*} \frac{45!}{3!1!1!7!1!1!4!1!5!2!2!1!4!2!6!1!1!2!}. \end{equation*}

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

By the Multinomial Theorem, the expansion of $(2x^3+y-z^2)^{100}$ has terms of the form

\begin{equation*} \binom{100}{k_1,k_2,k_3} (2x^3)^{k_1}y^{k_2}(-z^2)^{k_3} = \binom{100}{k_1,k_2,k_3} 2^{k_1}x^{3k_1}y^{k_2}(-1)^{k_3}z^{2k_3}. \end{equation*}

The $x^{99}y^{60}z^{14}$ arises when $k_1 = 33\text{,}$ $k_2=60\text{,}$ and $k_3=7\text{,}$ so it must have coefficient

\begin{equation*} -\binom{100}{33,60,7}2^{33}. \end{equation*}

For $x^{99}y^{61}z^{13}\text{,}$ the exponent on $z$ is odd, which cannot arise in the expansion of $(2x^3+y-z^2)^{100}\text{,}$ so the coefficient is $0\text{.}$