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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