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I came across something that confused me $$(2n)!=?$$ What does this mean: $$2!n!, \quad 2(n!)$$ or $$(2n)!=(2n)(2n-1)(2n-2)...n...(n-1)(n-2)...1$$ Which one is right?

The exercise is to show that $$(n+1)\bigg|\left(\begin{array}{c}2n\\n\end{array}\right)$$Then I thought of using the combination formula $\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!(n-k)!}$ to decrease my expression, but then I came across $$(2n)!$$

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Hint: You can verify by a computation that $$\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n+1}.$$

Detail: We have $$\frac{1}{n+1}\binom{2n}{n}=\frac{1}{n(n+1)}\frac{(2n)!}{(n-1)!n!}=\left(\frac{1}{n}-\frac{1}{n+1}\right) \frac{(2n)!}{(n-1)!n!} .$$ Now $$\frac{1}{n}\frac{(2n)!}{(n-1)!n!} =\binom{2n}{n}\qquad\text{and}\qquad \frac{1}{n+1}\frac{(2n)!}{(n-1)!n!} =\binom{2n}{n+1} .$$

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  • $\begingroup$ Yes, I agree that the right side is an integer, but could not follow your calculations, and understand what he had done. $\endgroup$ – marcelolpjunior Jul 24 '13 at 14:07
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You are correct: $$(2n)! = (2n)(2n-1)(2n-2)\cdots(3)(2)(1)$$

So, for example, if $n= 3$, then $(2n)! = 720$.

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  • $\begingroup$ Então, $(2n)!$ Can not be transformed into another expression with factorials? $\endgroup$ – marcelolpjunior Jul 24 '13 at 13:55
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Your expression for $(2n)!$ is correct. You can't easily split it down.

You might want to use $$\binom {2n}{n}=\frac {(2n)!}{n!n!}=\frac {n+1}{n}\frac{(2n)!}{(n+1)!(n-1)!}=\frac {n+1}n\binom {2n}{n+1}$$

noting that $n$ and $n+1$ are coprime, or consider $$\binom {2n}{n}-\binom {2n}{n+1}$$

These numbers are called Catalan numbers, and other proofs come from showing that they count discrete objects like the number of different (consistent) ways to insert pairs of brackets in an algebraic expression containing $n$ symbols [eg a(bc); (ab)c for three symbols].

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  • $\begingroup$ Correct, and interesting too, but in my statement, I did not understand as well as show using it. $\endgroup$ – marcelolpjunior Jul 24 '13 at 14:14
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If you have $(2n)!$, then is is indeed $$ (2n)! = (2n)(2n-1)(2m-2) \cdots 3\cdot 2\cdot 1. $$ So, you want to consider $$ \binom{2n}{n} = \frac{(2n)!}{n!(2n - n)!} = \frac{(2n)!}{(n!)^2} = \frac{(2n)(2n-1)\cdots (n+1)}{n!} $$ You want to show that $$ n+1 \mid \binom{2n}{n}. $$ That is, you want now to show that $n!\mid (2n)(2n-1)\cdots (n+2)$. You could try to do this using induction. For $n = 1$ (or $n=2$) this is obviously true. Then take it from there.

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  • $\begingroup$ Okay, I agree that the expression looks like this, but since this so will not help me in my demonstration. $\endgroup$ – marcelolpjunior Jul 24 '13 at 13:51
  • $\begingroup$ @marcelolpjunior: Let me add a bit more detail $\endgroup$ – Thomas Jul 24 '13 at 13:54
  • $\begingroup$ Have you seen the demonstration of Fermat's Little Theorem? $$p|a^p-a$$ If $p$ prime; As I recall, we used a demonstration that used a form of showing that some "numbers" fall combinations. $\endgroup$ – marcelolpjunior Jul 24 '13 at 14:04
  • $\begingroup$ @marcelolpjunior: I am not sure that I follow you... Andre's answer is better $\endgroup$ – Thomas Jul 24 '13 at 14:08
  • $\begingroup$ I am Brazilian, I use a translator, I do not understand English, sorry, I did not understand what you wrote last. $\endgroup$ – marcelolpjunior Jul 24 '13 at 14:10

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