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Stefan Kohl ♦
Stefan Kohl ♦
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You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$$$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor$$ where $x=\frac{m}{p^k},y=\frac{n}{p^k}$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$ where $x=\frac{m}{p^k},y=\frac{n}{p^k}$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor$$ where $x=\frac{m}{p^k},y=\frac{n}{p^k}$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

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user33772
user33772

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$ where $x=\lfloor\frac{m}{p^k}\rfloor,y=\lfloor\frac{n}{p^k}\rfloor$$x=\frac{m}{p^k},y=\frac{n}{p^k}$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$ where $x=\lfloor\frac{m}{p^k}\rfloor,y=\lfloor\frac{n}{p^k}\rfloor$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$ where $x=\frac{m}{p^k},y=\frac{n}{p^k}$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.

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user33772
user33772

You can show that $\frac{(2m)!(2n)!}{m!n!(m+n)!}$ is an integer by showing that each prime number $p$ divides the numerator at least as many times as it divides the denominator. For that it suffices to show that $$\lfloor\frac{2m}{p^k}\rfloor+\lfloor\frac{2n}{p^k}\rfloor\ge\lfloor\frac{m}{p^k}\rfloor+\lfloor\frac{n}{p^k}\rfloor+\lfloor\frac{m+n}{p^k}\rfloor$$ for natural $k$, i.e., that $$\lfloor2x\rfloor+\lfloor2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\\lfloor x+y\rfloor$$ where $x=\lfloor\frac{m}{p^k}\rfloor,y=\lfloor\frac{n}{p^k}\rfloor$. In fact, the latter inequality is easily seen to hold for all $x$ and $y$.