Question

In how many ways 10 coins can be distributed among 3 begars such that each begar must get atleast 1 coin ?

Answer 1

$\begin{array}{c}10\to coins\\ 3\to begars\end{array}|min\left(1\right)toeach$

$\to$ let's first give $1$ coin to each other begars to fulfill required condition.

Now we are left with $7\to coin$

To divide them in three parts two separators are needed to be inserted forex.

So, total no. of ways of dist $7$ coins $=$ Total no. of arrangements of these $7$ coins & $2$ sepanators

$=\frac{9!}{7!2!}=\frac{9\times 8}{\alpha }=36$