Question:
If $(n+1) !=12 \times[(n-1) !]$, find the value of $n .$
Solution:
To Find: Value of $n$
Given: $(n+1) !=12 \times[(n-1) !]$
Formula Used: $n !=(n) \times(n-1) \times(n-2) \times(n-3) \ldots \ldots \ldots .3 \times 2 \times 1$
Now, $(n+1) !=12 \times[(n-1) !]$
$\Rightarrow(n+1) \times(n) \times[(n-1) !]=12 \times[(n-1) !]$
$\Rightarrow(n+1) \times(n)=12$
$\Rightarrow n^{2}+n=12$
$\Rightarrow n^{2}+n-12=0$
$\Rightarrow(n-3)(n+4)=0$
$\Rightarrow n=3$ or, $n=-4$
But, n=-4 is not possible because in case of factorial (!) n cannot be negative.
Hence, n=3 is the correct answer