I know that $\sum_{k=0}^n{\left({n\choose{k}} \sum_{j=0}^{k-1}{{n-1}\choose{j}}\right)} = 2^{2n-2}$, but I can't figure out how to prove it. The sum came up in the context of a probability question, which can be solved via symmetry to show that the above is indeed $2^{2n-2}$. Any help would be greatly appreciated.
How to show$\sum_{k=0}^n{\left({n\choose{k}} \sum_{j=0}^{k-1}{{n-1}\choose{j}}\right)} = 2^{2n-2}$?
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summation
binomial-coefficients
2 Answers
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$$\sum_{k=0}^{n}\sum_{j=0}^{k-1}\binom{n}{k}\binom{n-1}{j} = \sum_{0\leq j < k \leq n}\binom{n}{k}\binom{n-1}{j} \tag{1}$$ can be written, by setting $d=k-j$, as $$ \sum_{d=1}^{n}\sum_{j=0}^{n-d}\binom{n}{j+d}\binom{n-1}{n-j-1}\stackrel{\text{Vandermonde}}{=}\sum_{d=1}^{n}\binom{2n-1}{n-d}=\sum_{d=0}^{n-1}\binom{2n-1}{d}\tag{2}$$ and the last sum is clearly half the sum $\sum_{d=0}^{2n-1}\binom{2n-1}{d} = 2^{2n-1}$, hence: $$\sum_{k=0}^{n}\sum_{j=0}^{k-1}\binom{n}{k}\binom{n-1}{j}=\color{red}{2^{2n-2}}\tag{3}$$ as wanted.
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We want to show that $\sum_{k=0}^n{n\choose{k}} \sum_{j=0}^{k-1}{{n-1}\choose{j}} = 2^{2n-2} $.
I almost do this, but there are two unexplained terms in my answer that prevent this.
So, I'll show what I have done and see is someone else can find what is wrong.
$\begin{array}\\ s(n) &=\sum_{k=0}^n{n\choose{k}} \sum_{j=0}^{k-1}{{n-1}\choose{j}}\\ &=\sum_{j=0}^{n-1}\sum_{k=j+1}^n{n\choose{k}} {{n-1}\choose{j}}\\ &=\sum_{j=0}^{n-1}{{n-1}\choose{j}}\sum_{k=j+1}^n{n\choose{k}} \\ &=\sum_{j=0}^{n-1}{{n-1}\choose{j}}(\sum_{k=0}^n{n\choose{k}}-\sum_{k=0}^j{n\choose{k}}) \\ &=\sum_{j=0}^{n-1}{{n-1}\choose{j}}(2^n-\sum_{k=0}^j{n\choose{k}}) \\ &=\sum_{j=0}^{n-1}{{n-1}\choose{j}}2^n-\sum_{j=0}^{n-1}{{n-1}\choose{j}}\sum_{k=0}^j{n\choose{k}}\\ &=2^{n-1}2^n-\sum_{j=0}^{n-1}\sum_{k=0}^j{{n-1}\choose{j}}{n\choose{k}} \\ &=2^{2n-1}-\sum_{k=0}^{n-1}\sum_{j=k}^{n-1}{{n-1}\choose{j}}{n\choose{k}} \\ &=2^{2n-1}-\sum_{k=0}^{n-1}{n\choose{k}} \sum_{j=0}^{k}{{n-1}\choose{j}}\\ &=2^{2n-1}-\sum_{k=0}^{n}{n\choose{k}} \sum_{j=0}^{k}{{n-1}\choose{j}} +{n\choose{n}} \sum_{j=0}^{n}{{n-1}\choose{j}}\\ &=2^{2n-1}-\sum_{k=0}^{n}{n\choose{k}} ({{n-1}\choose{k}}+\sum_{j=0}^{k-1}{{n-1}\choose{j}}) +2^{n-1}\\ &=2^{2n-1}-\sum_{k=0}^{n}{n\choose{k}} {{n-1}\choose{k}}-\sum_{k=0}^{n}{n\choose{k}}\sum_{j=0}^{k-1}{{n-1}\choose{j}} +2^{n-1}\\ &=2^{2n-1}-\sum_{k=0}^{n}{n\choose{k}} {{n-1}\choose{k}}-s(n) +2^{n-1}\\ \text{so}\\ 2s(n) &=2^{2n-1}-\sum_{k=0}^{n}{n\choose{k}} {{n-1}\choose{k}} +2^{n-1}\\ \end{array} $
At this point I am puzzled. If those last two terms were not there, we would have $s(n) = 2^{2n-2}$.
But they are there.
So either I have made a mistake or the original question is wrong.
I don't know which it is, so I'll leave it at this.