Discrete chocolates!

Generalizing, lets calculate the nymber of ways in which $n$ chocolates can be equally distributed among $n$ children.
So, each child should get $\dfrac{m}{n}$ chocolates.

First child can get chocolates in $^mC_{\frac{m}{n}}$ different ways. The The second child can get chocolates in $^{m-\frac{m}{n}}C_{\frac{m}{n}}$ different ways.
The third child can get chocolates in $^{m-2\frac{m}{n}}C_{\frac{m}{n}}$ different ways.
The fourth child can get chocolates in $^{m-3\frac{m}{n}}C_{\frac{m}{n}}$ different ways.
$\large \vdots$
The ${(n-1)}^{th}$ child can get chocolates in $^{m-(n-2)\frac{m}{n}}C_{\frac{m}{n}}$ different ways.
The $n^{th}$ child can get chocolates in $^{m-(n-1)\frac{m}{n}}C_{\frac{m}{n}}$ different ways.

So, the total number of ways in which the chocolates can be equally distributed
= $^mC_{\frac{m}{n}} × ^{m-\frac{m}{n}}C_{\frac{m}{n}} × ^{m-2\frac{m}{n}}C_{\frac{m}{n}} × \cdots × ^{m-(n-2)\frac{m}{n}}C_{\frac{m}{n}} × ^{m-(n-1)\frac{m}{n}}C_{\frac{m}{n}}\\ \\ = \dfrac{m!}{\left(\frac{m}{n}\right)! × \left(m - \frac{m}{n}\right)!} × \dfrac{\left(m - \frac{m}{n}\right)!}{\left(\frac{m}{n}\right)! × \left(m - 2\frac{m}{n}\right)! }× \dfrac{\left(m - \frac{2m}{n}\right)!}{\left(\frac{m}{n}\right)! × \left(m - \frac{3m}{n}\right)!} × \cdots × \dfrac{\left(m - (n-2)\frac{m}{n}\right)!}{\left(\frac{m}{n}\right)! × \left(m - (n-1)\frac{m}{n}\right)!} × \dfrac{\left(m - (n-1)\frac{m}{n}\right)!}{\left(\frac{m}{n}\right)! × \left(m - n\frac{m}{n}\right)!}\\ \\ = \color{#20A900}{\boxed{\dfrac{m!}{{\left(\frac{m}{n}\right)!}^n}}}$

So, plugging in $m = 12$ and $n= 3$ , we get, the number of ways of distributing the chocolates as:-
$\dfrac{12!}{{\left(\frac{12}{3}\right)}^3}\\ \\ = \dfrac{12!}{{\left(4!\right)}^3}\\ \\ = \color{#3D99F6}{\boxed{34650}}$