Number of ways to arrange three 1s, two 2s, two 3s and one 4?

Question

Number of ways to arrange three 1s, two 2s, two 3s and one 4? The order doesn't matter.

The direct answer from the book is $\frac{8!}{3!2!2!1!}$, but is there another way to do it?

Answer 1

Here is an iterative approach to find the number of ways to arrange three $1$s, two $2$s, two $3$s and one $4$:

• Three $1$s: We start with three $1$s: \begin{align*} 1\ 1\ 1\qquad\rightarrow\quad 1\ \mathrm{way}\quad \end{align*}
• Two $2$s: We have now $4$ positions where we can place two $2$s. We have $\binom{4}{1}$ ways to place two $2\mathrm{s}$ at one position and $\binom{4}{2}$ ways to place two $2$s at different positions: \begin{align*} \bullet\ 1\ \bullet\ 1\ \bullet\ 1\ \bullet \qquad\rightarrow\quad \binom{4}{1}+\binom{4}{2}\ \mathrm{ways} \end{align*}
• Two $3$s: To each of the valid arrangements consisting of three $1$s and two $2$s we have now $6$ positions where we can place two $3$s. We habe $\binom{6}{1}$ ways to place two $3$s at one position and $\binom{6}{2}$ ways to place two $3\mathrm{s}$ at different positions: \begin{align*} \bullet\ 1\ \bullet\ 1\ \bullet\ 2\ \bullet\ 1\ \bullet\ 2\ \bullet \qquad\rightarrow\quad \binom{6}{1}+\binom{6}{2}\ \mathrm{ways}\quad\qquad\quad \end{align*}
• One $4$: To each of the valid arrangements consisting of three $1$s, two $2$s and two $3$s we have now $\binom{8}{1}$ positions to place the number $4$. \begin{align*} \bullet\ 1\ \bullet\ 1 \bullet\ 3 \bullet\ 3\ \bullet\ 2\ \bullet\ 1\ \bullet\ 2\ \bullet \qquad\rightarrow\quad \binom{8}{1}\ \mathrm{ways}\qquad\qquad\qquad\qquad\qquad\quad \end{align*}

We conclude: The number of valid arrangements is \begin{align*} &1\cdot \left[\binom{4}{1}+\binom{4}{2}\right]\cdot\left[ \binom{6}{1}+\binom{6}{2}\right]\cdot\binom{8}{1}\\ &\qquad=\binom{5}{2}\binom{7}{2}8=10\cdot21\cdot 8\\ &\qquad\,\,\color{blue}{=1\,680} \end{align*} in accordance with $\frac{8!}{3!2!2!1!}=1\,680$.

Here we use the binomial identity $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$.