<s>[INST] Simplifying help: If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$. If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$.
So far, I have:
$\frac{n!}{3! (n-3)!} + \frac{(n+2)!}{3! (n-1)!}$
Then I simplified to:
$\frac{1}{6} (n-2)(n-1)(n) + \frac{1}{6} (n)(n+1)(n+2)$
Is this correct, and if so, what is the next move? It needs to equal $(n)_3$.
Thanks!
 [/INST] You are nearly ready.
Substitute $(n)_3 = n \cdot (n-1) \cdot (n-2)$, and you will see both sides are divisable by n. Thus, converting the result to a polinomial form, you will get an at most 2-grade equation, which will be soon very easy to solve.
$$\frac{1}{6}(n-2)(n-1)n + \frac{1}{6}n(n+1)(n+2) = n(n-1)(n-2)$$
Multiply with 6:
$$(n-2)(n-1)n + n(n+1)(n+2) = 6(n-2)(n-1)n$$
Minus $(n-2)(n-1)n$:
$$n(n+1)(n+2) = 5n(n-1)(n-2)$$
Divide both side with $n$:
$$(n+1)(n+2) = 5(n-1)(n-2)$$
$$n^2+3n+2 = 5(n^2-3n+2)$$
$$n^2+3n+2 = 5n^2-15n+10$$
$$4n^2-18n+8 = 0$$
$$2n^2-9n+4 = 0$$
$$n_{1,2}=\frac{9\pm \sqrt{81-4*2*4}}{4}=\frac{9\pm 7}{4} \in \{ \frac{1}{2}, 4\}$$
Only $\underline{\underline{4}}$ is integer, thus this is the only solution.
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