What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$? What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$?
How will I be able to find the answer to such questions? Is there any special technique to find the answer to such problems? Thank you.
 A: $18 = 2\cdot 3^2$. We can find the power of a small prime in a large factorial by successive division to find base divisibility, then divisibility by squares, etc. So the multiplicity of powers of $2$ in $50!$, $v_2(50!),$ is
$$ v_2(50!) = \left\lfloor\frac{50}{2}\right\rfloor + \left\lfloor\frac{50}{4}\right\rfloor + \left\lfloor\frac{50}{8}\right\rfloor + \cdots = 25+12+6+3+1 = 47$$
and similarly $v_2(25!)=22$, $v_3(50!)=16+5+1 = 22$ and $v_3(25!)=8+2 =10$,  so 
$$v_2\left(\frac{50!}{25!25!}\right) = 47-2\cdot22=3 \\
v_3\left(\frac{50!}{25!25!}\right) = 22-2\cdot 10=2 $$
and only $2$ available powers of $3$ means that $v_{18}\left(\frac{50!}{25!25!}\right)=1$ - the highest power of $18$ dividing the given expression is $18^1=18$.
A: HINT:Use the de polignac formula. As written first, answer was a little vague with many errors (I forgot that initially de polignac formula is used to find highest power of primes and 18 is not a prime). I m presenting a new one now. 
Since $18=2\times 3\times3$ (Prime factorisation). So, a better idea is just to find the highest power of $2,3,3$ in the numerator and denominator. Then  just subtract the number of powers of $2,3,3$ in denominator from the number of powers of $2,3,3$ in numerator. Arrange the remaining powers in such a manner so that they form highest possible power of $18$. That will be the answer.
