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? 
 A: To see that there is only one solution, let us generalize slightly and consider the function $f(x)=\binom{x}{k}$ where $x$ is an integer greater than or equal to $k$.  We wish to show that this is an increasing function.  
In theory, one could express $f(x)$ as a polynomial and use calculus.  However, we can take a combinatorial approach.
$f(x+1)$ is the number of ways to select $k$ elements from $\{1,2,\ldots,x+1\}$  We consider two types of subsets, the ones that contain $x+1$ and those that do not.  There are $f(x)$ subsets of the second type, and a positive number of the first type (as we are selecting $k-1$ elements from $\{1,2,\ldots,x+1\}$.
Therefore, $f(x+1)>f(x)$ and so $f(x)$ is an increasing function.  

I don't know that one can completely avoid guessing without resorting to calculus, but since the largest prime factor of $126*5!$ is $7$.  In particular, $11$, $13$, and $17$, are not factors, so you know that if $x>8$, then $x\geq 20$.  Additionally, $x\geq 5$.  If you verify that $8$ is too big, that gets you a small range to check.
A: Hints:
To find an estimate of the root, notice that the RHS is rather large and by $P(x)\approx x^5$, you can expect a root close to $\sqrt[5]{126\cdot5!}=6.85$, so you evaluate at $6$ and $7$.
The derivative of the product is a biquadratic polynomial, so that the extrema can be found and the impossiblity of other real roots confirmed.
A: Hint:
$$5!*126=9*8*7*6*5$$
$$(x+2)(x+1)(x)(x-1)(x-2)=9*8*7*6*5$$
A: 
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.
A: 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.
