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I'm suppose to proof the following with combinatorial proofs.

1)$$\sum_{i=0}^{n} {a+i \choose i} = {a+n+1 \choose n}$$

2)$$\sum_{i=0}^{n} i{n \choose i} = n2^{n-1}$$

3)$$\sum_{i=0}^{n} {n \choose i}^2 = {2n \choose n}$$

Any ideas how this is done ?

npisinp
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1 Answers1

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For the first one: enter image description here

For the second one check out this other post: Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$.

For the third just write the summand as $\binom{n}{i}\binom{n}{n-i}$. Imagine you have $2n$ students and you split the class into two chunks of $n$. You want to pick $n$ to go on a trip. You can choose first $0$ from one half, and the take other half, or one from one half and $n-1$ from the other half and so on. Or you could just straight up pick them without the partition in $\binom{2n}{n}$ ways.

  • Thanks, number 2 and 3 are useful. But I don't see any proof for the first one. – Franklin Apr 06 '14 at 19:35
  • @Franklin check this site. Q5 can prove your number 1. http://abc7434221math.blogspot.tw/2015/04/my-discrete-mathematics-midterm-exam.html#more – Vito Chou Apr 28 '15 at 07:04