Question

Find the value of x which satisfies the following equation.
$\frac{1}{10!}+\frac{1}{11!}=\frac{x}{12!}$

• A. 100
• B. 121
• C. 144
• D. 160

Answer 1

The correct option is C

144

$x!$ = $x(x-1)(x-2)......3.2.1$

$Given\frac{1}{10!}+\frac{1}{11!}=\frac{x}{12!}$

$\frac{12!}{10!}+\frac{12!}{11!}=x$

$So,\frac{12!}{10!}=\frac{12\times 11\times 10!}{10!}=132$

$Similarly,\frac{12!}{11!}=\frac{12\times 11!}{11!}=12$

So, the value of x = 132+12 = 144