A probability problem by Aly Ahmed

It is obvious that $6\leq x\leq 10$ .

Since $6\choose 6$ $=1$ ,

$7\choose 6$ $=7$ ,

$10\choose 10$ $=1$ ,

and $10\choose 9$ $=10$ , therefore $x=\boxed 8$ .

Given that

$\begin{aligned} \binom {10}x & = \frac {10!}{x!(10-x)!} \\ \frac {10!}{x!(10-x)!} & > 10 \\ x!(10-x)! & < 9! \\ \implies x & < 9 \end{aligned}$

$\begin{aligned} \binom x 6 & = \frac {x!}{6!(x-6)!} \\ \frac {x!}{6!(x-6)!} & > x \\ (x-1)! & > 6!(x-6)! \\ \implies x - 1 & > 6 \\ x & > 7 \end{aligned}$

Now $7 < x < 9 \implies x = \boxed 8$ .