Multinomial Theorem and Applications

Overview

The Multinomial Theorem generalises the Binomial Theorem.

The Binomial Theorem expands expressions of the form

(a+b)^n.

The Multinomial Theorem expands expressions of the form

(x_1+x_2+\cdots+x_m)^n.

It is useful for:

Expanding powers with more than two terms
Finding coefficients
Counting arrangements with repeated objects
Distributing choices among categories
Solving probability problems with more than two outcomes

Motivation

Consider

(x+y+z)^4.

A typical term in the expansion might be

x^2yz.

Where does this term come from?

We are multiplying

(x+y+z)(x+y+z)(x+y+z)(x+y+z).

To get $x^2yz$ , we have to choose:

$x$ from $2$ of the factors
$y$ from $1$ factor
$z$ from $1$ factor

The number of ways to decide which factors contribute each variable is

\frac{4!}{2!1!1!}.

So the coefficient of $x^2yz$ is

\frac{4!}{2!1!1!}=12.

This is the main idea behind the Multinomial Theorem.

Multinomial Coefficients

Suppose

k_1+k_2+\cdots+k_m=n.

The multinomial coefficient

\frac{n!}{k_1!k_2!\cdots k_m!}

counts the number of ways to arrange $n$ objects where there are:

$k_1$ objects of the first type
$k_2$ objects of the second type and so on

For example, the number of arrangements of the letters in

AABBC

\frac{5!}{2!2!1!}=30.

The Multinomial Theorem

For every nonnegative integer $n$ ,

(x_1+x_2+\cdots+x_m)^n = \sum \frac{n!}{k_1!k_2!\cdots k_m!} x_1^{k_1}x_2^{k_2}\cdots x_m^{k_m},

where the sum is taken over all nonnegative integers

k_1,k_2,\ldots,k_m

satisfying

k_1+k_2+\cdots+k_m=n.

Each term corresponds to choosing how many times each variable appears.

The Three Term Case

The most common version is

(x+y+z)^n.

The general term is

\frac{n!}{a!b!c!}x^ay^bz^c,

where

a+b+c=n.

This is especially useful for coefficient extraction.

Coefficient Extraction

To find a coefficient, determine how many times each term must be chosen.

Example 1

Find the coefficient of $x^2y^3z$ in
$(x+y+z)^6.$

The exponents are

2,3,1.

Their sum is

2+3+1=6,

so the term can appear.

The coefficient is

\frac{6!}{2!3!1!}=60.

Terms With Coefficients

If the expression contains coefficients, include them in the term.

Example 2

Find the coefficient of $x^2y z^2$ in $(2x-y+3z)^5$

To get

x^2yz^2,

we must choose:

$2x$ from $2$ factors
$-y$ from $1$ factor
$3z$ from $2$ factors

The multinomial coefficient is

\frac{5!}{2!1!2!}=30.

The contribution from the chosen terms is

2^2(-1)^1 3^2.

So the coefficient is

30(4)(-1)(9)=-1080.

Connection to Repeated Arrangements

The multinomial coefficient also counts arrangements with repeated objects.

Example 3

How many distinct arrangements are there of the letters in $MISSISSIPPI$ ?

The word has $11$ letters.

The counts are:

$M$ appears $1$ time
$I$ appears $4$ times
$S$ appears $4$ times
$P$ appears $2$ times

So the number of distinct arrangements is

\frac{11!}{1!4!4!2!}.

This is a multinomial coefficient.

Connection to Distribution

The multinomial coefficient also counts the number of ways to assign $n$ distinct objects into labelled groups of specified sizes.

Example 4

How many ways are there to split $10$ distinct students into labelled groups of sizes $2,3,5$ ?

The answer is

\frac{10!}{2!3!5!}.

This is the same coefficient that would appear in a multinomial expansion with exponents $2,3,5$ .

Counting All Outcomes

The Multinomial Theorem can also explain why

m^n

counts the number of ways to assign each of $n$ distinct objects to one of $m$ categories.

Set

x_1=x_2=\cdots=x_m=1.

Then

(x_1+x_2+\cdots+x_m)^n=m^n.

The multinomial expansion becomes the sum of all ways to split the $n$ choices among the $m$ categories.

Probability Applications

The multinomial distribution generalises binomial probability.

Suppose an experiment has $m$ possible outcomes with probabilities

p_1,p_2,\ldots,p_m.

If the experiment is repeated $n$ times independently, then the probability of getting outcome $1$ exactly $k_1$ times, outcome $2$ exactly $k_2$ times, and so on is

\frac{n!}{k_1!k_2!\cdots k_m!} p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m},

where

k_1+k_2+\cdots+k_m=n.

This is the probability version of the Multinomial Theorem.

Worked Probability Example

Example 5

A fair six-sided die is rolled $5$ times. What is the probability of rolling exactly two $1$ 's, one $2$ , and two numbers that are neither $1$ nor $2$ ?

There are three categories:

Rolling a $1$
Rolling a $2$
Rolling something else

Their probabilities are

\frac{1}{6}, \qquad \frac{1}{6}, \qquad \frac{4}{6}.

We want counts

2,1,2.

The probability is

\frac{5!}{2!1!2!} \left(\frac{1}{6}\right)^2 \left(\frac{1}{6}\right)^1 \left(\frac{4}{6}\right)^2.

The multinomial coefficient is

\frac{5!}{2!1!2!}=30.

So the probability is

30\left(\frac{1}{6}\right)^3\left(\frac{4}{6}\right)^2.

Multinomial Theorem and Binomial Theorem

The Binomial Theorem is the special case of the Multinomial Theorem with two terms.

When

m=2,

the Multinomial Theorem becomes

(x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k.

So binomial coefficients are just multinomial coefficients with two categories.

Worked Examples

Example 6

Find the coefficient of $a^2b^2c^3$ in
$(a+b+c)^7.$

The exponents are

2,2,3.

Their sum is

7.

So the coefficient is

\frac{7!}{2!2!3!}=210.

Example 7

Find the coefficient of $x^3y^2$ in
$(x+y+1)^5.$

To get

x^3y^2,

we choose:

$x$ from $3$ factors
$y$ from $2$ factors
$1$ from $0$ factors

The coefficient is

\frac{5!}{3!2!0!}=10.

Example 8

How many ways are there to arrange $4$ red balls, $3$ blue balls, and $2$ green balls in a row?

There are $9$ total balls.

The arrangements are counted by

\frac{9!}{4!3!2!}.

Key Contest Insight

The Multinomial Theorem is a counting formula disguised as algebra.

Each term in an expansion corresponds to choosing how many times each part of the sum is selected.

The same coefficient also counts arrangements of repeated objects and distributions into labelled groups of fixed sizes.

Strategy Tips

Identify the terms in the sum
Decide how many times each term must be chosen
Make sure the exponents add to the total power
Use the multinomial coefficient
Include numerical coefficients and signs from the terms
For repeated arrangements, divide by factorials of repeated counts
For probability, multiply the multinomial coefficient by the appropriate powers of the probabilities

Common Pitfalls

Forgetting that the exponents must add to $n$
Ignoring coefficients such as $2x$ or $-y$
Losing negative signs
Forgetting the $0!$ term when one variable is not chosen
Confusing multinomial coefficients with ordinary binomial coefficients
Treating repeated objects as distinct

Practice Problems

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