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

In summary, the expression ##^{2m}C_m## represents the number of ways to choose m elements from a set of 2m elements, which can be calculated using the formula for combinations. This formula is given by ##\dfrac{n!}{k!(n-k)!}##, where n is the total number of elements and k is the number of elements to choose. In this case, substituting n with 2m and k with m yields the equivalence ##^{2m}C_m = \dfrac{(2m)!}{m!(2m-m)!} = \dfrac{(2m)!}{m!m!}##, confirming that the two expressions are indeed equivalent.
  • #1
RChristenk
64
9
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!}##.
 
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  • #2
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*}
 
  • #3
fresh_42 said:
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!.
RChristenk said:
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!.
 
  • #4
fresh_42 said:
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!
 
  • #5
RChristenk said:
Yes you are correct I should've wrote ##(2m)!##. Thanks for your answer!
You should've written!
 
  • #6
martinbn said:
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.
 
  • #7
RChristenk said:
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".
 
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FAQ: Why is ##^{2m}C_m## equivalent to ##\dfrac{2m!}{m!m!}##?

What does ##^{2m}C_m## represent?

##^{2m}C_m## represents the number of ways to choose m items from a set of 2m distinct items. It is a binomial coefficient used in combinatorial mathematics to calculate combinations.

How is the formula for ##^{2m}C_m## derived?

The formula for ##^{2m}C_m## is derived from the general binomial coefficient formula ##^nC_k = \dfrac{n!}{k!(n-k)!}##. For ##^{2m}C_m##, n is 2m and k is m, so it becomes ##^{2m}C_m = \dfrac{2m!}{m!(2m-m)!} = \dfrac{2m!}{m!m!}##.

Why is the factorial used in the formula for combinations?

Factorials are used in the formula for combinations because they account for the number of ways to arrange a set of items. In the context of combinations, factorials help to calculate the total number of ways to choose a subset of items from a larger set, considering the order of selection does not matter.

Can you provide an example to illustrate ##^{2m}C_m = \dfrac{2m!}{m!m!}##?

Sure! Let's take m = 2. Then ##^{2m}C_m = ^4C_2##. According to the formula, this is ##\dfrac{4!}{2!2!} = \dfrac{24}{2 \cdot 2} = 6##. This means there are 6 ways to choose 2 items from a set of 4 distinct items.

How is this concept used in real-world applications?

This concept is widely used in probability theory, statistics, and various fields of science and engineering. For example, it is used in calculating probabilities in genetics, in designing experiments, and in analyzing algorithms in computer science.

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