Question

$\frac{72!}{36!36!}-1$ is divisible by

• A. $73$
• B. $72$
• C. $36$
• D. $37$

Answer 1

Answer: (A)

$73$

$\frac{72!}{36!36!}-1$

It can be written as $\frac{72\times 71\times \cdots \times 37}{36!}-1$

$\Rightarrow \frac{(73-1)\times (73-2)\times \cdots \times (73-36)}{36!}-1$

$\Rightarrow \frac{73\times \text{Some Number}}{36!}+\frac{1\cdot 2\cdots 36}{36!}-1$

$\Rightarrow \frac{73\times \text{Some Number}}{36!}$

It is now divisible by $73$

Hence, option A.