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How can I show that $(n!)^2$ divides $(2n)!$, where $n$ is a natural number?

So far I've noticed that we can rewrite $\dfrac{(2n)!}{(n)!^2}$ as a combination and we know that combinations are always natural, but is there an elegant way that makes more use of divisibility?

Sidd Singal
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Victor
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  • Do you mean $(2n)!/(n!)^2$? – Vincenzo Tibullo May 15 '14 at 20:34
  • Yes ,you're right , sorry ! – Victor May 15 '14 at 20:37
  • One can show, by induction or otherwise, that the product of $n$ consecutive integers is divisible by $n!$. Or else for our special case $(n!)^2$ and $(2n)!$, we can use the expression for the largest power of $p$ that divides $q!$. – André Nicolas May 15 '14 at 20:40
  • @mathguy Thanks for the link. Just noticed that one of the proofs there is false - still got 11 upvotes ... – user144248 May 15 '14 at 21:35
  • @Victor: I have posted an [answer to your question here.] (http://math.stackexchange.com/questions/796743/direct-proof-that-n-divides-n1n2-cdots2n) Sorry that it could not be done directly, but your question was incorrectly classified as a duplicate of a question about the evenness of the quotient. Of course the evenness implies integrality, but techniques to show integrality need not coincide with techniques to show evenness. – André Nicolas May 16 '14 at 18:45
  • Thank you very much for understanding that my problem needed a different approach in order to prove what's asked of us . Really nice proof you did there , thank you again , Mr. @AndréNicolas ! – Victor May 16 '14 at 20:59
  • You are welcome. – André Nicolas May 16 '14 at 22:07

2 Answers2

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Consider the fraction $$\frac{(2n)!}{(n!)^2}$$

It suffices to show that this is an integer. You may recognize the expression as the binomial coefficient

$$\binom{2n}{n}$$

All binomial coefficients are integers, and so, a fortiori, this one is as well. As for why all binomial coefficients are integers, the combinatorial interpretation of a binomial coefficient provides a good intuition.

If this isn't satisfying, I've included a bit of a more directly intuitive approach below.

Since $(2n)! = n! \cdot (n+1) \cdot (n+2) \cdots (2n)$, your original expression can be rewritten

$$\frac{n! \cdot (n+1) \cdot (n+2) \cdots (2n)}{n! \cdot n!}$$

$$\frac{(n+1) \cdot (n+2) \cdots (2n)}{n!}$$

Intuitively, why must every integer included in $n!$ have a multiple appear in the product $(n+1) \cdot (n+2) \cdots (2n)$?

Zubin Mukerjee
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    Sweet double answer, both with logic, and a great start to a rigorous proof – Asimov May 15 '14 at 20:54
  • But every integer included in $2\cdot 4$ also has a multiple appear in $4$, and yet $8\not\mid 4$. – user144248 May 15 '14 at 21:06
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    Honestly, I find you cannot simply say "every integer included in $n!$ has a multiple in $(n+1)\cdots (2n)$" and pretend that this completes the proof. Because every integer in, e.g., $1\cdot 2\cdot 4$ also has a multiple in $3\cdot 4\cdot 5$, but from this it doesn't follow that $8\mid 60$ (because it doesn't). – user144248 May 15 '14 at 21:53
  • I suppose the implicit criterion is that these multiples do not overlap - in your example, $2$ and $4$ both use $4$ as their multiple ... – Zubin Mukerjee May 15 '14 at 22:10
  • @ZubinMukerjee Well, in $1\cdot 2\cdot 3\cdot 4$ and $5\cdot 6\cdot 7\cdot 8$ both $2$ and $4$ use $8$ or both $2$ and $3$ use $6$. In other words, the multiples overlap here just as well. – user144248 May 16 '14 at 03:30
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    @ZubinMukerjee Btw. can you give a correct direct proof that $n!$ divides $(n+1)\cdots(2n)$? That the issue is subtle is illustrated by the false proof given in http://math.stackexchange.com/questions/78533/prove-that-2n-n2-is-even-if-n-is-a-positive-integer (by pedja and which got $11$ upvotes). Namely, in this proof the author claims that (using my above example) $5\cdot 6\cdot 7\cdot 8=2\cdot 2\cdot 4!\cdot 5\cdot 7$, but this is obviously false. – user144248 May 16 '14 at 03:37
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Put any prime $p$ and let $k$ and $\ell$ be the greatest positive integers such that $p^k|(n!)^2$ and $p^\ell|(2n)!$. Then using this \begin{align} k&=2\left(\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+\left\lfloor\frac n{p^3}\right\rfloor+\dots\right)\\ &=2\left\lfloor\frac np\right\rfloor+2\left\lfloor\frac n{p^2}\right\rfloor+2\left\lfloor\frac n{p^3}\right\rfloor+\dots\\ &\le \left\lfloor\frac {2n}p\right\rfloor+\left\lfloor\frac {2n}{p^2}\right\rfloor+\left\lfloor\frac {2n}{p^3}\right\rfloor+\dots\\ &=\ell. \end{align} It follows that $p^k|p^\ell$ which implies what you wanted.

Jlamprong
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