Verify that the equality

$\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$

when $n=3,r=2$, and then show that it always valid. (The sum is over all vectors of $r$ nonnegative integer values whose sum is $n$.)

Hint: How many different n letter sequences can be formed from the first $r$ letters of the alphabet? How many of them use letter $i$of the alphabet a total of $x_i$times for each $i=1,...,r$?

The given equality is $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$

and $n=3\text{and}r=2$

$x_1+x_2=n$, so the values of $x_1\text{and}{x}_2$are

localid="1649230670057" role="math" $\left(x_1=0,x_2=3\right),\left(x_1=1,x_2=2\right),\left(x_1=2,x_2=1\right),\left(x_1=3,x_2=0\right)$

Substituting the values in the equality $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$we get.

localid="1649230729929" role="math" $$\begin{gathered} \frac{3!}{0!3!}+\frac{3!}{1!2!}+\frac{3!}{2!1!}+\frac{3!}{3!0!}=2^3 \\ 1+3+3+!=8 \\ 8=8 \end{gathered}$$

Hence, it is proved thatlocalid="1649229296754" $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$.

The given equality is $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$

$n=3,r=3$

Consider, role="math" localid="1649229554269" $x_1+x_2+x_3=n$, so the values are

role="math" localid="1649230489627" $$\begin{gathered} \left(x_1=0,x_2=1,x_3=2\right),\left(x_1=1,x_2=2,x_3=0\right),\left(x_1=2,x_2=1,x_3=0\right),\left(x_1=3,x_2=0,x_3=0\right), \\ \left(x_1=0,x_2=2,x_3=1\right),\left(x_1=0,x_2=3,x_3=0\right),\left(x_1=0,x_2=1,x_3=2\right),\left(x_1=0,x_2=0,x_3=3\right), \\ \left(x_1=1,x_2=0,x_3=2\right),\left(x_1=2,x_2=0,x_3=1\right) \end{gathered}$$

Substituting the values in the equality $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$, we get

$$\begin{gathered} \frac{3!}{0!1!2!}+\frac{3!}{1!2!0!}+\frac{3!}{2!1!0!}+\frac{3!}{3!0!0!}+\frac{3!}{0!2!1!}+\frac{3!}{0!3!0!}+\frac{3!}{0!1!2!}+\frac{3!}{0!0!3!}+\frac{3!}{1!0!2!}+\frac{3!}{2!0!1!}=3^3 \\ 27=27 \end{gathered}$$

The given equality is $\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$and

$n=5,r=2$

Consider $x_1+x_2=n$, so the values are

$\left(x_1=0,x_2=5\right),\left(x_1=1,x_2=4\right),\left(x_1=2,x_2=3\right),\left(x_1=3,x_2=2\right),\left(x_1=4,x_2=1\right),\left(x_1=5,x_2=0\right)$

$$\begin{gathered} \frac{5!}{0!5!}+\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{3!2!}+\frac{5!}{4!1!}+\frac{5!}{5!0!}=2^5 \\ 32=32 \end{gathered}$$

Therefore, it is proved that$\sum _{x_1+....+x_r=n,x_i\ge 0}\frac{n!}{x_1!x_2!....x_r!}=r^n$.