16 different things are to be distributed among 3 persons A,B,C so that B gets 1 more than A and C get 2 more than B. In how many ways this can be done?

Answer 1

Let A get $x$ things, then B gets $x+1$ things and C gets $x+3$ things.

$\therefore \:x+x+1+x+3=16$

$\Rightarrow\:x=4$

$i.e.,$ A gets 4 things, B gets 5 things and C gets 7 things.

$\therefore$ The required no. of ways = $^{16}C_4\times ^{12}C_5\times ^7C_7$ ways.

$=\large\frac{16!}{4!.5!.7!}$ ways.