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Prove that ${{2n}\choose{n}}$ is even for $n \in \mathbb{N}$.

This one caught me off-guard when answering (or attempting to answer!) this for a student today. I tried this approach:

$${{2n}\choose{n}}=\frac{(2n)!}{n!n!}=\frac{(2n)(2n-1)(2n-2)\dots (n+1)}{n!}$$

and recognized that rearranging the numerator as $$(2n)(2(n-1))(2(n-2)) \ldots (2n-1)(2n-3) \ldots(n+1)$$

can help, but I don't know, roughly speaking, how far the chain on the left of the dots above goes.

Darrin
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4 Answers4

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Combinatorial proof: From a set of $2n$ chocolates, choose $n$ to eat. However, these choices come in pairs -- the $n$ I didn't choose I could have equally well chosen. Hence, all ${2n\choose n}$ "menus" can be paired off, so there must be an even number of them.

vadim123
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  • That I can see, but I think the method would require more of a direct proof, if possible. – Darrin Mar 12 '14 at 03:12
  • (The problem is listed in Book Of Proof by Hammack, Chapter 4, number 23, in the Direct Proof section) – Darrin Mar 12 '14 at 03:13
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    @Darrin I haven't read the book, but this is a direct proof as opposed to a proof by contradiction (or even a proof by induction). – Mark S. Mar 12 '14 at 03:16
  • This is clear and intuitive. However, one would have to take care of the case for which a pair contains twice the same "menu" (n=0), and for which the statement is wrong. – Wok Mar 12 '14 at 14:06
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    @Wok, $\mathbb{N}$ is commonly understood to exclude $0$. – vadim123 Mar 12 '14 at 14:11
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    Not in my country. In France, we call ℕ* the set excluding 0. – Wok Mar 12 '14 at 14:13
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    While it's not standard, it is common to have $\mathbb{N}$ not include zero. See this earlier discussion. – vadim123 Mar 12 '14 at 14:19
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    With Wiki, "there is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century." In the context, this is ambiguous since "For natural numbers (taken to include 0) n and k", the binomial coef is defined – Wok Mar 12 '14 at 14:19
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    It is good policy, when a question is ambiguously stated, to choose the interpretation under which it is correct. – vadim123 Mar 12 '14 at 14:24
  • I see your point, yet I believe it is a shame to have ambiguous statements in math. – Wok Mar 12 '14 at 14:26
  • @Darrin What do you mean by direct proof? By the usual definition (as mentioned by @MarkS), this qualifies. – Kyle Strand Mar 12 '14 at 22:17
  • @KyleStrand Perhaps he means a proof by mathematical operations, rather than an intuitive explanation of why it has to be true. – Brilliand Mar 12 '14 at 23:51
  • vadim123 clearly labeled it a "combinatorial proof" which means the standard intuitive counting proofs we often give in combinatorics. – Applied Squared Mathematician Mar 13 '14 at 00:49
  • By direct proof I would definitely say the one provided here qualifies, but for the purposes of my student (who is new to mathematics) and the spirit of the book he is using, the demonstration I marked as the answer would be the leap I was looking for. I will, though, present BOTH solutions here to him when we meet again tomorrow. – Darrin Mar 13 '14 at 04:28
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$$\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$$ while $$\binom{2n-1}{n-1} = \binom{2n-1}{(2n-1)-(n-1)} = \binom{2n-1}{n}$$ so $$\binom{2n}{n} = 2 \times \binom{2n-1}{n} $$

Fei Gao
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I think @vadim123 has given an excellent proof, but if the OP wants to use only algebraic methods, here goes.

Lemma. If $n$ is a positive integer and $n\le 2^k$ and $$s=\Bigl\lfloor\frac{n}{2}\Bigr\rfloor+\Bigl\lfloor\frac{n}{4}\Bigr\rfloor+\Bigl\lfloor\frac{n}{8}\Bigr\rfloor+\cdots+\Bigl\lfloor\frac{n}{2^k}\Bigr\rfloor\ ,$$ then $2^s$ is a factor of $n!$ and $2^{s+1}$ is not.

Furthermore, $$\eqalign{s &\le\frac{n}{2}+\frac{n}{4}+\frac{n}{8}+\cdots+\frac{n}{2^k}\cr &<\frac{n}{2}+\frac{n}{4}+\frac{n}{8}+\cdots+\frac{n}{2^k}+\cdots\cr &=n\ ,\cr}$$ so $n!$ is not a multiple of $2^n$.

Now write $$\eqalign{\binom{2n}{n} &=\frac{(2n)(2n-2)\cdots(2)}{n!}\frac{(2n-1)(2n-3)\cdots(1)}{n!}\cr &=2^n\,\frac{(2n-1)(2n-3)\cdots(1)}{n!}\ .\cr}$$ The expression has $n$ factors of $2$ in the numerator, and from the above results, not all of them will be cancelled out by the $n!$ in the denominator. So the expression is even.

David
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The above answers are already proofs, however I think mine is a simpler one.

Just using the definition of the binomial form:

$$\binom{2n}{n} = \frac{2n*(2n-1)\ *\cdots*(2n-(n-1))}{n!}=$$ $$=\frac{2n*(2n-1)*\cdots*(2n-(n-1))}{n*(n-1)!}=$$ $$=\frac{2n}{n}*\frac{(2n-1)*\cdots*(2n-(n-1))}{(n-1)!}=$$ $$=2*\frac{(2n-1)*\cdots*(2n-(n-1))}{(n-1)!}=$$ $$=2*\binom{2n-1}{n-1}$$ Two times a whole number will be even.

p1100i
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