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? 

Use brute force approach, We can easily find x=7 is the solution.
And x=7 is the only one solution. Because:
There's no solution for x<3, because $\binom{x+2}{5}=0$ for x<3
for x>=3:
$\binom{x0+2}{5} < \binom{x+2}{5} < \binom{x1+2}{5}$ for every x1 > x > x0 >=3; and x,x1,x0 is real number
so 
$\binom{x0+2}{5} < \binom{7+2}{5} = 126 < \binom{x1+2}{5}$ for every x1 > 7 > x0 >=3; and x1,x0 is real number
Conclusion: x=7 is the unique answer.
