# How many different ways are there of arranging the letters in the word BARBARIAN?

Jan 28, 2016

$15120$

#### Explanation:

There are $\left(\begin{matrix}9 \\ 2\end{matrix}\right)$ ways to place the $\text{B}$s.

From the remaining $7$ spots, there are $\left(\begin{matrix}7 \\ 3\end{matrix}\right)$ ways to place the $\text{A}$s.

From the remaining $4$ spots, there are $\left(\begin{matrix}4 \\ 2\end{matrix}\right)$ ways to place the $\text{R}$s.

In the remaining $2$ spots, there are 2! ways to arrange the $\text{I}$ and the $\text{N}$

Thus the total possible arrangements is

((9),(2))((7),(3))((4),(2))2! = (9!)/(7!2!)(7!)/(4!3!)(4!)/(2!2!)2! =(9!)/(3!2!2!) = 15120

Note that we could also write this as
(9!)/(3!2!2!1!1!)

with the factorials in the denominator matching the "groups" of letters ($3$ $\text{A}$s, $2$ $\text{B}$s, $2$ $\text{R}$s, $1$ $\text{I}$, $1$ $\text{N}$). Looking at how the cancellation worked when we did our initial multiplication, we can see how it will work out this way in general, and so as a shorthand, we can write this using the multinomial coeffficient

$\left(\begin{matrix}\null & \null & 9 & \null & \null \\ 3 & 2 & 2 & 1 & 1\end{matrix}\right)$