Proving $\sum_{a_1+a_2+a_3=n}{n\choose a_1, a_2, a_3}=3^n$ I need some hints where to start to prove that
$$\sum_{a_1+a_2+a_3=n}{n\choose a_1, a_2, a_3}=3^n$$
 A: Let $X$ a set with cardinal $n$ then the number of maps $X\to\{1,2,3\}$ is $3^n$ and each map is defined if we associate the elements $x_1,\ldots,x_{a_1}$ of $X$ to $1$ and  $y_{1},\ldots,y_{a_2}$ of $X$ to $2$ and  $z_1,\ldots,z_{a_3}$ of $X$ to $3$ where 
$$X=\{x_1,\ldots,x_{a_1},y_{1},\ldots,y_{a_2},z_1,\ldots,z_{a_3}\}$$
so we have
$$\sum_{a_1+a_2+a_3=n}{n\choose a_1,a_2,a_3}=3^n$$
A: Hint:  Use the multinomial theorem by substitution
A: Hint: When $a_1+a_2+a_3=n$,
$$
\begin{align}
\binom{n}{a_1,a_2,a_3}
&=\frac{n!}{a_1!a_2!(n-a_1-a_2)!}\\[6pt]
&=\frac{n!}{a_1!(n-a_1)!}\frac{(n-a_1)!}{a_2!(n-a_1-a_2)!}\\[6pt]
&=\binom{n}{a_1}\binom{n-a_1}{a_2}
\end{align}
$$
Sum in $a_2$ then sum in $a_1$.
A: Hint What is the coefficient of $x^{a_1}y^{a_2}z^{a_3}$ in $(x+y+z)^n$.
What happens if you set $x=y=z=1$?
A: This is a repetition of the answer of Sami Ben Roundhane, so if you feel like upvoting, upvote his instead.
There are $3^n$ words of length $n$ over the alphabet $\{a,b,c\}$. We count the words another way. 
For any non-negative integers $p$, $q$, and $r$ such that $p+q+r=n$, there are $\binom{n}{p,q,r}$ words that have exactly  $p$ $a$'s, $q$ $b$'s, and $r$ $c$'s. For the locations of the $p$ $a$'s, $q$ $b$'s, and $r$ $c$'s can be chosen in $\binom{n}{p,q,r}$ ways. Summing over all $p$, $q$, and $r$ such that $p+q+r=n$ gives us a count of all the words. 
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{a_{1} + a_{2} + a_{3}\ =\ n}{n \choose a_{1},\ a_{2},\ a_3}
     = 3^{n}:\  {\large ?}}$

\begin{align}&\color{#00f}{\large
\sum_{a_{1} + a_{2} + a_{3}\ =\ n}{n \choose a_{1},\ a_{2},\ a_3}}
=\sum_{a_{1}=0}^{\infty}\sum_{a_{2}=0}^{\infty}\sum_{a_{3}=0}^{\infty}
{n! \over a_{1}!\ a_{2}!\ a_{3}!}\,\delta_{a_{1} + a_{2} + a_{3},n}
\\[3mm]&=\sum_{a_{1}=0}^{\infty}\sum_{a_{2}=0}^{\infty}\sum_{a_{3}=0}^{\infty}
{n! \over a_{1}!\ a_{2}!\ a_{3}!}\int_{\verts{z}\ =\ 1}
{1 \over z^{-a_{1} - a_{2} - a_{3} + n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z}\ =\ 1}{n! \over z^{n + 1}}\,\pars{%
\sum_{a = 0}^{\infty}{z^{a} \over a!}}^{3}\,{\dd z \over 2\pi\ic}
=\int_{\verts{z}\ =\ 1}n!\,{\expo{3z} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
=\sum_{k = 0}^{\infty}n!\ {3^{k} \over k!}\
\overbrace{\int_{\verts{z}\ =\ 1}\,{z^{k} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}}
^{\ds{\delta_{kn}}}
\\[3mm]&=\color{#00f}{\LARGE 3^{n}}
\end{align}

