Skip to main content
Logo image

Elementary Foundations: An introduction to topics in discrete mathematics

Jeremy Sylvestre

Section 23.2 Multinomial Coefficients

Similarly to the proof of the Binomial Theorem, write
\begin{gather} (x + y + z)^n = (x + y + z) (x + y + z) \dotsm (x + y + z) \text{,}\tag{✶} \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 \(\combinationalt{n}{i}\) ways to choose \(i\) factors from the right-hand side of (✶) from which to take \(x\text{,}\) then \(\combinationalt{n - i}{j}\) 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(\choosefuncformula{n}{i}{(n-i)}\right) \:\: \left(\choosefuncformula{(n-i)}{j}{(n-i-j)}\right) = \frac{n!}{i! \, j! \, k!}\text{.} \end{equation*}
Use the Binomial Theorem on \(\bbrac{x + (y + z)}^n\text{,}\) then again on \((y + z)^k\) for each term \(\combinationalt{n}{k} x^{n - k} (y + z)^k\text{.}\) (This would be very tedious!)

Worked Example 23.2.2. 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 = \bbrac{(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*} \amp (2 x + y - 3 z)^3 \\ \amp = 8 x^3 + y^3 - 27 z^3 + 12 x^2y - 36 x^2 z \\ \amp \phantom{=} + 6 x y^2 - 9 y^2 z - 54 x z^2 - 36 x y z \text{.} \end{align*}

Worked Example 23.2.3. 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*} \frac{14!}{5! \, 2! \, 7!} = \frac{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.
Use the same generalized FOIL method argument as in the Binomial and Trinomial Theorem proofs, and simplify the product of combination formulas obtained.

Worked Example 23.2.5. 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 = \bbrac{(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*} \frac{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 \(1,088,640\text{.}\)
multinomial coefficient
a number appearing as a coefficient in the expansion of \((x_1 + x_2 + \dotsb + x_m)^n\)
\(\binom{n}{i_1,i_2,\dotsc,i_m}\)
the coefficient on the term \(x_1^{i_1} x_2^{i_2} \dotsm x_m^{i_m}\) in the expansion of \((x_1 + x_2 + \dotsm + x_m)^n\text{,}\) where the exponents \(i_1, i_2, \dotsc, i_m\) must sum to \(n\)

Note 23.2.6.

  • The Multinomial Theorem tells us \(\displaystyle \binom{n}{i_1,i_2,\dotsc,i_m} = \frac{n!}{i_1! \, i_2! \, \dotsm \, i_m!} \text{.}\)
  • 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} = \choosefuncformula{n}{i_1}{i_2} = \choosefuncformula{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.
<Prev^TopNext>