How to solve a binomial for $n$

Question

How do you solve the equation:

$$ \binom{n}{3}=4\binom{n}{2} $$

I've tried multiple times I'm just not sure what I'm supposed to do to be honest.

What I did try doing was the combinatorial theorem, and re-arranging to get n, but I didnt get a solution. So I'm wondering how I would go about solving for n.

Answer 1

You can just write the definitions on each side and solve for $n$:

$$\dfrac {n!}{3!(n-3)!}=4\dfrac {n!}{2!(n-3)!}$$

$$\iff \dfrac {1}{6(n-3)!}=\dfrac {2}{(n-2)!}$$

$$\iff \dfrac {(n-2)!}{(n-3)!}=12$$

$$\iff (n-2)=12$$

$$\iff n=14$$

Answer 2

If you write out these expressions in terms of their definitions, you are greeted with simply a polynomial equation:

$${n\choose3} = 4{n\choose2} \implies \frac{n(n-1)(n-2)}{3!}=\frac{4n(n-1)}{2!}.$$

From here, it should be very easy to extract such a value for $n$.