Trinomial Coefficients

Recall from the Binomial Coefficients page that if $n$ and $k$ are nonnegative integers that satisfy $0 \leq k \leq n$ then the binomial coefficient $\binom{n}{k}$ is defined to be $\displaystyle{\binom{n}{k} = \frac{n!}{k! (n - k)!}}$. We saw that these binomial coefficients played an important rule in deriving The Binomial Theorem.

Note that in the formula for the binomial coefficients, we have that the sum of the numbers in the denominator equals the numerator, that is $n = k + (n - k)$. If we instead let $r_1 = k$ and $r_2 = n - k$, then $n = r_1 + r_2$ to get that $\binom{n}{k} = \binom{n!}{r_1! \cdot r_2!}$. We can instead use the notation $\binom{n}{r_1, r_2}$ to mean the same thing.

We now have a starting point to extend the idea of the binomial coefficients to what are called trinomial coefficients and we will see the relationship these trinomial coefficients have with The Trinomial Theorem later on.

Definition: If $r_1$, $r_2$, and $r_3$ are nonnegative integers and $n = r_1 + r_2 + r_3$ then the Trinomial Coefficient $\displaystyle{\binom{n}{r_1, r_2, r_3}}$ is defined to be $\displaystyle{\binom{n}{r_1, r_2, r_3} = \frac{n!}{r_1! \cdot r_2! \cdot r_3!}}$.

For example, let's compute the value of $\binom{10}{2, 3, 5}$. We get:

(1)

\begin{align} \quad \binom{10}{2, 3, 5} = \frac{10!}{2! \cdot 3! \cdot 5!} = 2520 \end{align}

Of course, it should be rather obvious that if $r_1 + r_2 + r_3 = n$ then by the commutativity of multiplication:

(2)

\begin{align} \quad \binom{n}{r_1, r_2, r_3} = \binom{n}{r_1, r_3, r_2} = \binom{n}{r_2, r_1, r_3} = \binom{n}{r_2, r_3, r_1} = \binom{n}{r_3, r_1, r_2} = \binom{n}{r_3, r_2, r_1} \end{align}

In other words, for $r_1 + r_2 + r_3 = n$ there will be $3!$ ways to represent the same trinomial coefficient.