Prove the identity $ \sum_{i=0}^n {n \choose i} i = n2^{n-1} $ using combinatorial proof.
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kingnewbie
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You are trying to explain this by counting a given set in two different ways, like one of the previous questions you posted about $\binom{52}{13}$ and I answered? – Sarvesh Ravichandran Iyer Sep 14 '16 at 10:04
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That was a different question. I know it may seem the same, but that was actually asking to prove it with another identity; this question is asking for a proof which is using combinatorial methods. – kingnewbie Sep 14 '16 at 10:29
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You are not searching for popular questions (regarding these binomial identities) which have been adequately answered. If you have any specific query then post it, but your 3 questions on these identities are really duplicates. – StubbornAtom Sep 14 '16 at 11:27
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I have looked at the other solutions posted to the same question. The reason why I posted this was to actually understand what 'counting arguments' means because none of the other questions explicitly said this. – kingnewbie Sep 14 '16 at 11:28
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4Have you checked this question and the questions which are linked there? For example, this one. And this questions looks like direct duplicate. (And again, looking at the linked questions might be useful, too.) – Martin Sleziak Sep 14 '16 at 11:34
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Hint. Consider the following counting problem:
How many ways do you have of choosing a non-empty subset of $S_n:=\{1,2,\dots,n\}$ and then an element of this subset?
You can choose a subset of $i$ elements in $\binom{n}{i}$ ways (note that $\sum_{i=0}^n \binom{n}{i}$ gives you all the subsets of $S_n$ that is $2^n$) then you pick one of the $i$ element of this subset in $i$ ways. Hence the total number is $$\sum_{i=1}^n {n \choose i} i,$$ that is the left-hand side of your identity.
What about the right-hand side?
Another approach: note that $$\sum_{i=1}^n {n \choose i} i=\sum_{i=1}^n \frac{n}{i}\cdot {n-1 \choose i-1} i=n\cdot\sum_{i=1}^n {n-1 \choose i-1}=n\sum_{j=0}^{n-1} \cdot {n-1 \choose j}=n2^{n-1}.$$.
Robert Z
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$\binom ni$ is the number of ways to pick $i$ items out of $n$, and in your sum there is an extra factor $i$. So we can count all the ways to draw $i$ items and paint one of them red. This is the same as painting one item red and picking $i-1$ white ones in all possible ways.
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I don't see a better way to describe it than as a "counting argument". Unless you would prefer "paint shop" ? – Sep 14 '16 at 10:33