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I have played around with this question for a bit, but my answer doesn't feel right. I'm not happy with how I converted the binomial. it feels a bit hand-wavy. Is there any way I can make this proof better?

Let's say that we have two teams that are both size $n$. Let's call one team the Blue team and the other team Red. We are going to send a mixture of Blue and Red players to an event (they can also send no one). There are $\binom{2n}{a}$ ways we can select players, but if we want to only send an even sum of players then $a$ must be even. Lets rewrite it as $\binom{2n}{2k}$. Considering all possible $k$-values tells us that the total ways to send an even sum of randomly composed Blue and Red players is $$\sum_{k=0}^n \binom{2n}{2k}$$ Since both teams have $n$ players, we know that there are $2^{2n}$ subgroups that could be sent. But again we only want to consider even groups so we divide by 2 to get rid of all odd groups. Thus we have

$$2^{2n}/2 = 2^{2n-1}$$

Both methods give the same result.

rohitt
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WaterDrop
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3 Answers3

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No, your approach is incorrect. When you say "we divide by $2$ and get rid of all odd groups" you assert that the number of even groups is equal to the number of odd groups, in other words $$\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=0}^{n-1}\binom{2n}{2k+1}$$ which is equivalent to $$2\cdot\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=0}^{n}\binom{2n}{2k}+\sum_{k=0}^{n-1}\binom{2n}{2k+1}=2^{2n}$$

You essentially proved that $$\sum_{k=0}^{n}\binom{2n}{2k}=2^{2n-1}\text{ assuming }\sum_{k=0}^{n}\binom{2n}{2k}=2^{2n-1}$$

As @Trevor Gunn pointed out, Pascal's indetity can be used to get the result.

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Using Pascal's identity:

$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

you can write this as

$$ \binom{2n-1}{0} + \binom{2n-1}{1} + \binom{2n-1}{2} + \dots + \binom{2n-1}{2n-1} = 2^{2n-1}. $$

This proof can be written purely combinatorialy if you wish, just use the combinatorial proofs of Pascal's identity and $2^n = \sum_k \binom{n}{k}$.

You can also get a bijection between even/odd subsets of $\{1,\dots,2n-1\}$ by mapping a subset to its complement (the complement of an even subset is odd and vice versa). This is enough to prove the identity as you've seen and is explained more succinctly in Vlad's answer.

Trevor Gunn
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I find the algebraic approach crystal clear: $$ 2^{2n} = 2^{2n}+ 0 = (1+1)^{2n} + (1-1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} + \sum_{k=0}^{2n} (-1)^k\binom{2n}{k} = 2 \sum_{k=0\\k \text{ even}}^{2n} \binom{2n}{k} $$

lhf
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