Binomial coefficients inequation problem Can anyone help me solve this:
$$5\binom{13}{x} &lt; \binom{x + 2}{4}$$
After turning it to factorial I don't know what to do nothing seems to cancel out. $x$ is a positive integer.
I end up with this when trying to solve it:
$$\frac{13!}{x(x+1)(13 - x)!} &lt; \frac{(x+2)!}{5!}$$
Edit: I found solutions in a book with this task, says solutions are {11, 12, 13} I've plugged them in and they seem to work. But can anyone explain how it was solved?
$x&gt;10$.  I got this simply by plugging in all the values of x from 2 to 13.

Here is the Mathematica code and output I wrote:

Do[Print[5*13!/(x!*(13 - x)!) - (x + 2)!/(4!*(x + 2 - 4)!)], {x, 2, 13}]

389
1425
3560
6400
8510
8454
6225
3245
935
-325
-936
-1360
$$5\binom{13}{x} &lt; \binom{x + 2}{4}$$
can be seen as:
\begin{align}
5 \, \frac{13!}{x! \, (13-x)!} &amp;&lt; \frac{(x+2)!}{4! \, (x-2)!} \\
5! \, 13! \, (x-2)! &amp;&lt; x! \, (13-x)! \, (x+2)! \\  
5! \, 13! \, x! &amp;&lt; x(x-1) \, x! \, (x+2)! \, (13-x)! \\
5! \, 13! &amp;&lt; x(x-1) \, (x+2)! \, (13-x)! 
\end{align}
From this can be stated that, for $x$ being an integer, $2 \leq x \leq 13$ for allowable values. Note that by inspection the right-hand binomial limits the value of $x$ to $x \leq 13$. The right-hand binomial provides the lower limit $x \geq 2$. By utilizing the allowable values of $x$ it is seen that the values $x \in \{11, 12, 13\}$ are valid solutions.