MHB Ordered pair (x,y): x choose y = 2020

[juantheron]

Finding all natural number ordered pair $(x,y)$ for which $\displaystyle \binom{x}{y} = 2020.$

 

[Olinguito]

My attempt (may be wrong).

Spoiler

Given a binomial coefficient $\displaystyle\binom xy$, we can make two observations:

• For fixed $x$, $\displaystyle\binom xy$ increases as $y$ increases from $0$ to $y=\left\lfloor\dfrac x2\right\rfloor$.
• For fixed $y=1,\ldots,\left\lfloor\dfrac x2\right\rfloor$, $\displaystyle\binom xy$ increases as $x$ increases.

Now $\displaystyle\binom xy=2020$ $\implies$ $x!=2020\cdot y!\cdot(x-y)!$. So the prime $101$ divides $x!$ since it divides $2020$. Thus we must have $x\ge101$.

But $\displaystyle\binom{101}2\ =\ 5050\ >\ 2020$.

It follows from the two observations above (and the fact that $\displaystyle\binom xy=\binom x{x-y}$) that $\displaystyle\binom xy>2020$ for all $x\ge101$ and $y=2,3,\ldots,x-2$.

Hence the only integers $x,y$ such that $\displaystyle\binom xy=2020$ are $(x,y)=(2020,1),(2020,2019)$.

 

[juantheron]

Thanks https://mathhelpboards.com/members/olinguito/ My solution is almost same as you.