23.2: Multinomial Coefficients

Theorem \(\PageIndex{1}\): Trinomial Theorem.

The expansion of the trinomial \((x + y + z)^n\) is the sum of all possible products

\begin{equation*} \dfrac{n!}{i! \, j!\, k!}\,x^i y^j z^k, \end{equation*}
where \(0 \le i,j,k \le n\) such that \(i + j + k = n\text{.}\)

Proof Idea.

Similarly to the proof of the Binomial Theorem, write

\begin{gather} (x + y + z)^n = (x + y + z) (x + y + z) \cdots (x + y + z) \text{,}\label{equation-multinomial-trinomial}\tag{\(\star\)} \end{gather}
with \(n\) factors. To expand this out, we generalize the FOIL method: from each factor, choose either \(x\text{,}\) \(y\text{,}\) or \(z\text{,}\) then multiply all your choices together. For any such product, the powers on \(x\text{,}\) \(y\text{,}\) and \(z\) must sum to \(n\text{.}\) To get the final expansion, add the results of all possible such products.

But we can collect terms that have the same exponent on each of \(x\text{,}\) \(y\text{,}\) and \(z\text{.}\) How many ways can we form a specific term \(x^i y^j z^k\text{,}\) for \(0 \le i,j,k \le n\) such that \(i + j + k = n\text{?}\) We have \(C^n_i\) ways to choose \(i\) factors from the right-hand side of (\(\star\)) from which to take \(x\text{,}\) then \(C^{n-i}_j\) ways to choose \(j\) factors from which to take \(y\text{.}\) But now from all remaining factors we must choose \(z\text{,}\) and there is only \(1\) way to do this. So the coefficient on \(x^i y^j z^k\) is

\begin{equation*} \binom{n}{i} \binom{n-i}{j} = \left(\dfrac{n!}{i! (n-i)!}\right) \:\: \left(\dfrac{(n-i)!}{j! (n-i-j)!}\right) = \dfrac{n!}{i! \, j! \, k!}\text{.} \end{equation*}

Alternative proof idea.

Use the Binomial Theorem on \((x + (y + z))^n\text{,}\) then again on \((y + z)^k\) for each term \(C^n_k x^{n - k} (y + z)^k\text{.}\) (This would be very tedious!)

Example \(\PageIndex{1}\): Expanding a trinomial.

Determine the terms in the expansion of \((2 x + y - 3 z)^3\text{.}\)

Solution

First, rewrite

\begin{equation*} (2 x + y - 3 z)^3 = ((2 x) + y + (-3 z))^3 \text{.} \end{equation*}
So the terms in the expansion involve products

\begin{equation*} (2 x)^i y^j (-3 z)^k \text{.} \end{equation*}
We need to account for all triples of exponents \(i, j, k\) that sum to \(3\text{.}\)

\(i\)	\(j\)	\(k\)	\(n! \over i! \, j! \, k! \)	term	simplified
\(3\)	\(0\)	\(0\)	\(1\)	\((2 x)^3 \)	\(8 x^3 \)
\(0\)	\(3\)	\(0\)	\(1\)	\(y^3 \)	\(y^3 \)
\(0\)	\(0\)	\(3\)	\(1\)	\((-3 x)^3 \)	\(-27 z^3 \)
\(2\)	\(1\)	\(0\)	\(3\)	\(3 (2 x)^2 y \)	\(12 x^2y \)
\(2\)	\(0\)	\(1\)	\(3\)	\(3 (2 x)^2 (-3 z) \)	\(-36 x^2 z \)
\(1\)	\(2\)	\(0\)	\(3\)	\(3 (2 x) y^2 \)	\(6 x y^2 \)
\(0\)	\(2\)	\(1\)	\(3\)	\(3 y^2 (-3 z) \)	\(-9 y^2 z \)
\(1\)	\(0\)	\(2\)	\(3\)	\(3 (2 x) (-3 z)^2 \)	\(-54 x z^2 \)
\(0\)	\(1\)	\(2\)	\(3\)	\(3 y (-3 z)^2 \)	\(-9 y z^2 \)
\(1\)	\(1\)	\(1\)	\(3!\)	\(6 (2 x) y (-3 z) \)	\(-36 x y z \)

Collecting this together, we have

\begin{align*} & (2 x + y - 3 z)^3 \\ & = 8 x^3 + y^3 - 27 z^3 + 12 x^2y - 36 x^2 z \\ & + 6 x y^2 - 9 y^2 z - 54 x z^2 - 36 x y z \text{.} \end{align*}

Example \(\PageIndex{2}\): Determining a specific coefficient in a trinomial expansion.

Determine the coefficient on \(x^5 y^2 z^7\) in the expansion of \((x + y + z)^{14}\text{.}\)

Solution

Here we don't have any extra contributions to the coefficient from constants inside the trinomial, so using \(n=14\text{,}\) \(i = 5\text{,}\) \(j = 2\text{,}\) \(k = 7\text{,}\) the coefficient is simply

\begin{equation*} \dfrac{14!}{5! \, 2! \, 7!} = \dfrac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8} {5 \cdot 4 \cdot 3 \cdot 2 \cdot 2} = 14 \cdot 13 \cdot 11 \cdot 9 \cdot 4 = 72,072\text{.} \end{equation*}

The pattern of the Binomial Theorem and Trinomial Theorem continues.

Theorem \(\PageIndex{2}\): Multinomial Theorem.

The expansion of \((x_1 + x_2 + \cdots + x_m)^n\) is the sum of all possible products

\begin{equation*} \dfrac{n!}{i_1! \, i_2! \, \cdots \, i_m!} \, x_1^{i_1} x_2^{i_2} \cdots x_m^{i_m} \text{,} \end{equation*}
where the exponents \(i_1, i_2, \ldots, i_n\) sum to \(n\text{.}\)

Proof Idea.

Use the same generalized FOIL method argument as in the Binomial and Trinomial Theorem proofs, and simplify the product of combination formulas obtained.

Example \(\PageIndex{3}\): Determining a specific coefficient in a multinomial expansion.

Determine the coefficient on \(x^2 y z^6\) in the expansion of \((3 x + 2 y + z^2 + 6)^8\text{.}\)

Solution

Rewriting

\begin{equation*} (3 x + 2 y + z^2 + 6)^8 = ((3 x) + (2 y) + (z^2) + 6)^8 \text{,} \end{equation*}
we see that the four terms in this multinomial are

\begin{equation*} 3 x, \quad 2 y, \quad z^2, \quad 6 \text{.} \end{equation*}
So what we really want to know is the total coefficient on the term involving

\begin{equation*} (3 x)^2 (2 y)^1 (z^2)^3 6^2 \text{.} \end{equation*}
The Multinomial Theorem tells us that there will be

\begin{equation*} \dfrac{8!}{2! \, 1! \, 3! \, 2!} = 1,680 \end{equation*}
such terms in the expansion of the multinomial. Therefore, we obtain the term

\begin{equation*} (1,680) (3 x)^2 (2 y)^1 (z^2)^3 6^2 = (1,088,640) x^2 y z^6 \end{equation*}
with a total coefficient of \(1,088,640\text{.}\)

Definition: Multinomial Coefficient

a number appearing as a coefficient in the expansion of \((x_1 + x_2 + \cdots + x_m)^n\)

Definition: \(\binom{n}{i_1,i_2,\ldots,i_m}\)

the coefficient on the term \(x_1^{i_1} x_2^{i_2} \cdots x_m^{i_m}\) in the expansion of \((x_1 + x_2 + \cdots + x_m)^n\text{,}\) where the exponents \(i_1, i_2, \ldots, i_m\) must sum to \(n\)

Note

• The Multinomial Theorem tells us \(\displaystyle \binom{n}{i_1,i_2,\ldots,i_m} = \dfrac{n!}{i_1! \, i_2! \, \cdots \, i_m!} \text{.}\)
• In the case of a binomial expansion \((x_1 + x_2)^n\text{,}\) the term \(x_1^{i_1} x_2^{i_2}\) must have \(i_1 + i_2 = n\text{,}\) or \(i_2 = n - i_1\text{.}\) The Multinomial Theorem tells us that the coefficient on this term is

\begin{equation*} \binom{n}{i_1,i_2} = \dfrac{n!}{i_1!i_2!} = \dfrac{n!}{ i_1! (n - i_1)!} = \binom{n}{i_1}. \end{equation*}
Therefore, in the case \(m=2\text{,}\) the Multinomial Theorem reduces to the Binomial Theorem.