Browse Questions

# If $\large\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$ find $x$

$\begin{array}{1 1}(A)\;8\\(B)\;32\\(C)\;64\\(D)\;128\end{array}$

Toolbox:
• $n!=n(n-1)(n-2)........(3)(2)(1)$
$\large\frac{1}{6!}=\frac{1}{6\times 5\times 4\times 3\times 2\times 1}$
$\large\frac{1}{7!}=\frac{1}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
$\large\frac{1}{6!}+\frac{1}{7!}=\frac{1}{6\times 5\times 4\times 3\times 2\times 1}+\frac{1}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
$\Rightarrow \large\frac{7+1}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
$\Rightarrow \large\frac{8}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
Multiply by 8 in both numerator and denominator
$\Rightarrow \large\frac{8\times 8}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}$
$\Rightarrow \large\frac{8\times 8}{8!}$
Given :
$\large\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$
$\large\frac{64}{8!}=\frac{x}{8!}$
$\Rightarrow 64$
Hence (C) is the correct answer.