Question

$12$ balls are distributed among three boxes find the probability that first box will contain $3$ balls.

Answer 1

$12$ balls are distributed among $3$ boxes

So, total ways $=3^{12}$

1st box containing $3$ balls

${}^{12}{C}_3\times 2^9=\frac{12\times 11\times 10\times 9!}{3\times 2\times 1\times 9!}\times 2^9=110\times 2^{10}$

So, probability $=\frac{110\times 2^{10}}{3^{12}}$