23.2: Multinomial Coefficients
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!)
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*}
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.
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.
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 \(1,088,640\text{.}\)
a number appearing as a coefficient in the expansion of \((x_1 + x_2 + \cdots + x_m)^n\)
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 \(n\)
- 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
.