Saudi Arabian 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 $x \geq 179$.

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 $y \leq \left\lfloor \frac{x}{2} \right\rfloor$. Then
$$\binom{x}{1} < \binom{x}{2} < \ldots < \binom{x}{\left\lfloor \frac{x}{2} \right\rfloor}.$$
If $y=1$, then we get $x=1432$. If $y \geq 2$, then we have
$$\binom{x}{y} \geq \binom{x}{2} = \frac{x(x-1)}{2} \geq \frac{179 \cdot (179-1)}{2} > 1432,$$
not possible.