<s>[INST] Solve $\binom{x+2}{5} = 126$ for $x$ How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one solution exists? 
 [/INST] 
Use:

*

*$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$

*$$a!=\prod_{p=1}^{a}p=a\times(a-1)\times(a-2)\times\dots\times3\times2\times1$$


So, we get:
$$\binom{x+2}{5}=126\Longleftrightarrow\frac{(x+2)!}{5!((x+2)-5)!}=126\Longleftrightarrow$$
$$\frac{(x+2)!}{120(x-3)!}=126\Longleftrightarrow\frac{(x+2)!}{(x-3)!}=15120$$
Expanding $\frac{(x+2)!}{(x-3)!}$:
$$\frac{(x+2)!}{(x-3)!}=x(x-2)(x-1)(x+2)(x+2)=x(x^4-5x^2+4)$$
So, we get:
$$x(x^4-5x^2+4)=15120\Longleftrightarrow(x-7)(x^4+7x^3+44x^2+308x+2160)=0$$
So, we will get $7$ as a solution.
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