Solving for $n$

Question

$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2} = 22$$

I'm trying to solve this equation for $n$.

If $$\binom{n}{0} = 1$$

Then we have that

$$1+\binom{n}{1}+\binom{n}{2} = 22 \implies \binom{n}{1}+\binom{n}{2} = 21$$

However, I'm stuck there. Could you help me out?

Regards

Answer 1

Note that$$\binom n0+\binom n1+\binom n2=1+n+\frac{n(n-1)}2.$$Therefore, your equation is a quadratic one.

Answer 2

Since for $n=7$ we get ${7\choose 2} = 21$ the $n$ can not be $\geq 8$. So you could try to solve it by inspection for all $n\leq 7$.