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As the title says...

We are asked to show that $$\sum_{i=1}^{n}\binom{n}{i}\binom{n}{i-1}=\binom{2n}{n-1}$$

I tried with induction, but that seems to never work with these kind of questions. We need to understand what we are counting on the left side. The right side is obvious, we are counting the ways to choose $n-1$ elements out of a set with $2n$ elements. But what are we counting on the left side? Give a hint

Alexander Gruber
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Oria Gruber
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2 Answers2

4

Hints:

  • ${2n \choose n-1}={2n \choose n+1}\\$
  • ${n \choose i-1}={n \choose n-i+1}$
  • ${n \choose i}{n \choose n-i+1}$ - amount of ways to choose $i$ elements from the group of $n$ elements and choose $n-i+1$ elements from remaining $n$ elements.
Norbert
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  • This will help me solve it algebraically? – Oria Gruber Jan 20 '14 at 21:53
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    No this will help to solve combinatorically – Norbert Jan 20 '14 at 21:55
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    if the republican and democrat parties have each $n$ members, and we want $n+1$ members to be in the parliament, then we have $n+1$ members to choose from $2n$. Or we could choose $i$ democrats and then choose the rest of the candidates ($n+1-i$) from the republicans, and the summation of all those options with all $i$s is the number of ways we can choose. – Oria Gruber Jan 20 '14 at 22:09
  • @OriaGruber Good example! – Norbert Jan 20 '14 at 22:11
  • It is enough to change only the first term: $\binom ni=\binom n{n-i}$. – Berci Jan 20 '14 at 22:23
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Hint: Note that the coefficient of $x^{n-1}$ in the expansion of $(1+x)^{2n}$ is the RHS.

Now $(1+x)^{2n}=(1+x)^n(x+1)^n$. The coefficient of $x^{n-1}$ in the expansion of $(1+x)^n(x+1)^n$ is the LHS (after a bit of trivial algebraic manipulation.)

voldemort
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