Question

If $\left(n+1\right)!=12\times (n-1)!thenn=$

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (A)

3

We have,

$\left(n+1\right)!=12\times (n-1)!$

$(n+1)\times n\times (n-1)!=12\times (n-1)!$

$(n+1)\times n=12$

$n^2+n-12=0$

$n^2+4n-3n-12=0$

$(n+4)(n-3)=0$

$(n+4)\ne 0$

So,

$n-3=0$

$n=3$

Hence, this is the answer.