If $n\ge2$, Prove $\binom{2n}{3}$ is even.

Question

Any help would be appreciated. I can see it's true from pascal's triangle, and I've tried messing around with pascal's identity and the binomial theorem to prove it, but I'm just not making any headway.

Answer 1

$${2n\choose 3}=\frac{(2n)(2n-1)(2n-2)}{6}=\frac{n(2n-1)(n-1)\cdot 2}{3}$$ Note that the numerator is a multiple of 2, while the denominator is not, hence the result must be a multiple of 2.

In fact, since one of $n, n-1$ must be even, the number ${2n\choose 3}$ must be a multiple of $4$, not just $2$.

Answer 2

Combinatorial arguments are good, too.

Imagine that you have $n$ couples, for a total of $2n$ people. Let $\mathscr{T}$ be the collection of all $\binom{2n}3$ sets of $3$ of these people. For each $T\in\mathscr{T}$ let $T'$ be the member of $\mathscr{T}$ obtained by trading each member of $T$ for that person’s spouse. Thus, if $T$ does not contain a couple, then $T\cap T'=\varnothing$, while if $T$ does contain a couple, $T'$ contains the same couple and the spouse of the third member of $T$. Note that in every case $T\ne T'$, and $(T')'=T$.

$$\big\{\{T,T'\}:T\in\mathscr{T}\big\}$$

is then a partition of $\mathscr{T}$ into two-element sets, so $\binom{2n}3=|\mathscr{T}|$ is even.

Answer 3

Here's an approach not using factorials:

Take $2n$ objects and separate them into two groups of $n$. You're interested in the number of ways you can choose $3$ of them. For any such choice, you'll have more from one pile than the other. By symmetry, there'll be the same number of choices that "favor" each pile over the other, so in total there will be an even number of choices.

This argument works to prove the evenness of $2n\choose k$ for any odd number $k$.

Answer 4

$\displaystyle\ \dfrac{(2n)(2n\!-\!1)(2n\!-\!2)}{3\cdot 2\cdot 1} \, =\, \dfrac{\color{#c0f}4}{\color{#0a0}{2n\!+\!1}}\dfrac{(\color{#0a0}{2n\!+\!1})(2n)(2n\!-\!1)(2n\!-\!2)}{\color{#c0f}4\cdot 3\cdot 2\cdot 1}\, =\, \dfrac{\color{#c00}4}{2n\!+\!1}{2n\!+\!1\choose 4}\ $ is $\,\rm\color{#c00}{even}$

Remark $\ $ The same method shows $\ k\!+\!1 \mid {\large {\,n\,\choose k}}\, $ if $\ \gcd(k\!+\!1,n\!+\!1)=1,\,$ e.g.

$$ a^{\large k}\, {\Large \mid}\ {an\choose a^{\large k}\!-1}\qquad\quad$$

Your example is simply the special case $\ a=2 = k.$

Answer 5

In the spirit of various other answers, let $S=\{\color{red}1,\color{blue}1,\color{red}2,\color{blue}2\dots, \color{red}n,\color{blue}n\}$. If $T$ is a $3$-element subset of $S$, one color predominates among its elements, and the set $T'$, obtained by changing the color of every number in $T$, is a different $3$-element subset of $S$. (The other color predominates.) The $3$-element subsets of $S$ can thus be collected into disjoint pairs, each subset paired with its “color opposite,” so there must be an even number of these subsets.

Answer 6

(for $n\gt2$) we have the numerator (after the initial cancellation of $(2n-3)!)$ is at least divisible by $8$, however the denominator can at most be divisible by $2$.

Answer 7

For any size-$3$ subset of $\{1,2,\dots, 2n\}$, trade each element $a$ for $2n+1-a$. This is an involution; that just means that if you do it again you get back to the same thing. Therefore, we can simply pair up the subsets according to this rule.

We might be worried that something gets paired up with itself; this of course would ruin the whole endeavor. But notice that each element in $\{n+1,n+2,\dots, 2n\}$ is sent to an element in $\{1,2,\dots, n\}$ and the opposite is also true. Since there must be an odd number of elements in the subsets we are counting, there must be an odd number of their elements in the bottom half or the top half, but not both. After performing the rule above, which half has the odd number switches, so the set rule could not possibly send any set to itself.

(This proof has the advantage of generalizing very easily to show that $\binom nk$ is even whenever $n$ is even and $k$ is odd. It's the same as Brian's, although he found a more imaginative scenario.)

Answer 8

$\binom{2n}{3}=\dfrac{(2n)!}{3!\cdot (2n-3)!}= \dfrac{2n(2n-1)(2n-2)}{6}$. Use induction.

Answer 9

By definition:

$\binom{2n}{3} = \frac{(2n)!}{(2n-3)! 3!} = \frac{(2n)(2n-1)(2n-2)}{3 \times 2} = \frac{(n)(2n-1)(n-1)\times 2}{3}$

Since the binomial coefficient is always an integer, we conclude that at least one of the terms $n, 2n-1 $ or $n-1$ must be divisible by 3 (which must be provable), which leaves us with a number multiplied by 2 in the numerator, which is even. QED.