If $\binom{n+2}{5}=12\binom{n}3$ what is n.

Question

If $\binom{n+2}{5}=12\binom{n}3$ what is n.

I have tried taking the long way of breaking down what (n+2C5) and 12(nC3) are. However, it has led me to a wrong answer.

The tip is:

Use the property: $\binom nk=\frac{n}k\binom{n-1}{k-1}.$

Answer 1

$${n+2\choose 5} = 12{n\choose 3}$$

So $${(n+2)!\over 5!\cdot (n-3)!}=12{n!\over 3!(n-3)!}$$

So $${(n+2)(n+1)n!\over 5\cdot 4\cdot 3! (n-3)!}=12{n!\over 3!(n-3)!}$$

So $${(n+2)(n+1)\over 20} =12$$

Can you finish now?

Answer 2

$${n+2\choose 5} = 12{n\choose 3}\implies (n+2)(n+1)=240 $$

The positive answer which is acceptable is $n=14$