Question

The number of ways to give $16$ different things to three persons $A,B,C$ so that $B$ gets one more than $A$ and $C$ gets two more than $B$, is

• A. $\frac{16!}{4!5!7!}$
• B. $4!5!7!$
• C. $\frac{16!}{3!5!8!}$
• D. None of these

Answer 1

The correct option is A $\frac{16!}{4!5!7!}$

Given,
$B=A+1$
$C=2+B=3+A$

$A+B+C=16$
$3+A+A+1+A=16$
$3A=12$
$A=4,B=5,C=7$

So,number of ways $=\frac{16!}{4!5!7!}$