Saudi Arabia Mathematical Competitions

We have $1432=2^3\cdot 179$. The equation is equivalent to $$\frac{x!}{y!(x-y)!}=2^3\cdot 179$$ or $y!(x-y)!\cdot 2^3\cdot 179=x!$. It follows $179\mid x!$, hence $x\ge 179$.

It is clear that $(x,y)=(1432,1),(1432,1431)$ are solutions. We shall prove that there are no other solutions. Because of the symmetry of the binomial coefficients, we can assume that $y\le \lfloor \frac{x}{2}\rfloor$. Then $$\left(\genfrac{}{}{0ex}{}{x}{1}\right)<\left(\genfrac{}{}{0ex}{}{x}{2}\right)<\dots <\left(\genfrac{}{}{0ex}{}{x}{\lfloor \frac{x}{2}\rfloor }\right).$$ If $y=1$, then we get $x=1432$. If $y\ge 2$, then we have $$\left(\genfrac{}{}{0ex}{}{x}{y}\right)\ge \left(\genfrac{}{}{0ex}{}{x}{2}\right)=\frac{x(x-1)}{2}\ge \frac{179\cdot (179-1)}{2}>1432,$$ not possible.