Why is ##^{2m}C_m## equivalent to ##\dfrac{2m!}{m!m!}##?

[RChristenk]

Homework Statement

Why is $^{2m}C_m$ equivalent to $\dfrac{2m!}{m!m!}$?

Relevant Equations

Elementary combination principles

By definition, $^nC_r=\dfrac{n(n-1)(n-2)...(n-r+1)}{r!}$. This can be simplified to $^nC_r=\dfrac{n!}{r!(n-r)!}$, which leads to $^{2m}C_m=\dfrac{2m!}{m!m!}$.

But I can't see how from the original equation $^{2m}C_m=\dfrac{(2m)(2m-1)(2m-2)...(m+1)}{m!}$ is equivalent to $\dfrac{2m!}{m!m!}$.

 

[fresh_42]

What do you mean? Your formula is incorrect. It has to be $(2m)!$ and not $2m!$
\begin{align*}
^{2m}C_m&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)}{m!}\\&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m}{m!\cdot m}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)}{m!\cdot m\cdot (m-1)}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-2)}{m!\cdot m\cdot (m-1)\cdot (m-2)}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)}{m!\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)}\\
&\vdots \\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)\cdots 2\cdot 1}{m!\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)\cdots 2\cdot 1}\\
&=\dfrac{(2m)!}{m!m!}
\end{align*}

 

[Mark44]

What do you mean? Your formula is incorrect. It has to be (2m)! and not 2m!

I'm pretty sure that (2m)! is what this poster meant, but wrote incorrectly as 2m!.

By definition, $^nC_r=\dfrac{n(n-1)(n-2)...(n-r+1)}{r!}$. This can be simplified to $^nC_r=\dfrac{n!}{r!(n-r)!}$, which leads to $^{2m}C_m=\dfrac{2m!}{m!m!}$.

But I can't see how from the original equation $^{2m}C_m=\dfrac{(2m)(2m-1)(2m-2)...(m+1)}{m!}$ is equivalent to $\dfrac{2m!}{m!m!}$.

Multiply numerator and denominator by m!.

 

[RChristenk]

What do you mean? Your formula is incorrect. It has to be $(2m)!$ and not $2m!$
\begin{align*}
^{2m}C_m&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)}{m!}\\&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m}{m!\cdot m}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)}{m!\cdot m\cdot (m-1)}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-2)}{m!\cdot m\cdot (m-1)\cdot (m-2)}\\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)}{m!\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)}\\
&\vdots \\
&=\dfrac{(2m)(2m-1)(2m-2)...(m+1)\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)\cdots 2\cdot 1}{m!\cdot m\cdot (m-1)\cdot (m-3)\cdot (m-3)\cdots 2\cdot 1}\\
&=\dfrac{(2m)!}{m!m!}
\end{align*}

Yes you are correct I should've wrote $(2m)!$. Thanks for your answer!

 

[martinbn]

Yes you are correct I should've wrote $(2m)!$. Thanks for your answer!

You should've written!

 

[RChristenk]

You should've written!

I'm sorry I didn't write $(2m)!$. I will in the future do my earnest to double-check my work in the future. Please don't take offense at my negligence.

 

[martinbn]

I'm sorry I didn't write $(2m)!$. I will in the future do my earnest to double-check my work in the future. Please don't take offense at my negligence.

You misuderstood. I was pedantic about "I should've wrote".