Multinomial Theorem: Let \(k \geq 3\) and \(n \geq 2.\) Let \(A(k) = \{1,2,\dots,k\}.\) Define the set of \(k\)-tuples \[J(k,n) = \left\{(a_1, a_2, \dots, a_k) : \left(\forall i \in A_k, a_i \geq 0\right) \wedge \left(\sum_{i \in A_i} a_i = n\right)\right\}\] Then \[\left(\sum_{i=1}^k x_i\right)^n = \sum_{(a_1, \dots, a_k) \in J(k,n)} {n \choose a_1,a_2,\dots,a_k}\prod_{i=1}^k x_i^{a_i}\] where \[{n \choose a_1,a_2,\dots,a_k} = \frac{n!}{a_1!a_2!\dots a_k!}\]

We will not prove the multinomial theorem, but we will show a couple of examples.

Example: Use the multinomial theorem to compute \((x+y+z)^2.\)

There are \(3\) terms in the sum, and the sum is raised to the power \(2.\) In this case, we use \(J_{3,2}\) which is all the ways to add \(3\) non-negative integers to get \(2.\) \[J_{3,2} = \{(2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1)\}\] Using the multinomial theorem, we get \begin{align} & (x + y + z)^2 \\ & = {2 \choose 2,0,0}x^2y^0z^0 + {2 \choose 0,2,0}x^0y^2z^0 + {2 \choose 0,0,2}x^0y^0z^2 + {2 \choose 1,1,0}x^1y^1z^0 + {2 \choose 1,0,1}x^1y^0z^1 + {2 \choose 0,1,1}x^0y^1z^1 \\ & = \frac{2!}{2!0!0!}x^2 + \frac{2!}{0!2!0!}y^2 + \frac{2!}{2!0!2!}z^2 + \frac{2!}{1!1!0!}xy + \frac{2!}{1!0!1!}xz + \frac{2!}{0!1!1!}yz \\ & = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \end{align}

Example: Use the multinomial theorem to find the coefficient of \(xy^2\) in the polynomial \((2x-y+3z-2)^6.\)

There are \(4\) terms in the sum, and the sum is raised to the power \(6.\) In this case, we use \(J_{4,6}\) which is all the ways to add \(4\) non-negative integers to get \(6.\)

We are only interested in the coefficient of the term \(xy^2,\) which will be the term where the summands are raised to the following powers: \((2x)^1(-y)^2(3z)^0(-2)^3.\) The element of \(J_{4,6}\) we are concerned with is \((1,2,0,3).\) The coefficient is \[{6 \choose 1,2,0,3} = \frac{6!}{1!2!0!3!} = 60\] Therefore, the term with \(xy^2\) will be \[60(2x)^1(-y)^2(3z)^0(-2)^3 = -960xy^2.\] The coefficient of \(xy^2\) is \(-960.\)