Question:
if$\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$, find the value of $x$
Solution:
To Find: Value of n
Given: $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$
Formula Used: $n !=(n) \times(n-1) \times(n-2) \times(n-3)$ $3 \times 2 \times 1$
Now, $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$
$\Rightarrow \frac{1}{24}+\frac{1}{120}=\frac{x}{720}(4 !=24,5 !=120)$
$\Rightarrow \frac{5+1}{120}=\frac{x}{720}$
$\Rightarrow \frac{6}{120}=\frac{x}{720}$
$\Rightarrow x=36$