natural number values of $(x,y)$ in binomial coefficients

Question

Find natural number $(x,y)$ in $\displaystyle \frac{x!}{y!(x-y)!}=2019$

what i try

$\displaystyle \frac{x!}{y!(x-y)!}=2019=3\cdot 673$

pairs are $(2019,1),(2019,2018)$

How to find other natural values of $(x,y)$ help me please

Answer 1

An important fact: Numbers in same row of Pascal triangle not coprime

Another important fact is that each row of the pascal triangle strictly increases until the middle point of symmetry. This can be easily shown by induction with the addition property of the triangle. I.e, if $k$th row is strictly increasing until middle, then the sum of two consecutive terms is also strictly increasing until the middle point implying the next row is strictly increasing until the middle point.

Two formulas to be used are:

($i$) ${x \choose y} = {x-1\choose y-1} + {x-1\choose y}$.

($ii$) ${x \choose y} = {x-1\choose y-1}\times{x\over y}$

Case $1$: ${x-1\choose y-1}$ is a multiple of $673$. This means it is either $673$ or $1346$, the latter case implying ${x-1\choose y} = 673$. However a prime number appears in only the second index of a row in the pascal triangle since a number in the middle of the pascal triangle is larger than the second number in the same row and is not co-prime with that number. This would mean either ${x-1\choose y-1}$ or ${x-1\choose y}$ is the second number of the $673$th row. We write the first three terms of that row: $1, 673, 226128$ and sees clearly non of the two consecutive ones add up to $2019$.

Case $2$: ${x-1\choose y-1}$ is not a multiple of $673$. Since it times $x\over y$ becomes a multilple of $673$, this means $x$ is a multiple of $673$. So $x$ is one of $673, 1346, 2019$. Since ${x \choose y}$ lies on $x$th row of the pascal triangle, we simply write out the three rows' first several elements. $673$ is already done. $1346$th row: $1, 1346, 905185, ...$ and strictly increasing until middle point so clearly no 2019 appears. As a result $x=2019$ is the only number left. The second number in the row gives $2019$ and its mirror. Since the row is then strictly increasing no other entry in this row gives $2019$.