Binomial coefficients

\(\text{Binomial coefficients} { n\choose k}, n,k \in \mathbb{N}_0, k \leq n, \text{are defined as}\)

(ni)=n!i!(ni)!{n \choose i}=\frac{n!}{i!(n-i)!}.

They satisfy(ni)+(ni1)=(n+1i) for i>0\text{They satisfy}{n \choose i}+{n \choose i-1}={n+1 \choose i} \text{ for } i > 0

and also(n0)+(n1)++(nn)=2n,\text{and also} {n \choose 0}+{n\choose 1}+\cdots+{n \choose n}=2^{n},

(n0)(n1)++(1)n(nn)=0,{n \choose 0}-{n\choose 1}+\cdots+(-1)^{n}{n \choose n}=0,

(n+mk)=i=0k(ni)(mki).{n+m \choose k}=\sum\limits_{i=0}^k {n \choose i} {m \choose k-i}.

How do I prove that\text{How do I prove that}

(n+mk)=i=0k(ni)(mki)?{n+m \choose k}=\sum\limits_{i=0}^k {n \choose i} {m \choose k-i}?

(Edit: This is also known as the Vandermonde’s Identity.)\text{(Edit: This is also known as the Vandermonde's Identity.)}

Help would be greatly appreciated. (I came across this in a book)\text{Help would be greatly appreciated. (I came across this in a book)}


By the way, I used LaTeX to type the whole note :)\text{By the way, I used LaTeX to type the whole note :)}

Note by Victor Loh
7 years ago

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Here's a combinatorial proof.


From a group of m+nm+n students consisting of nn boys and mm girls, how many ways are there to form a team of kk students?

Answer 1:


Answer 2:

If that team has ii boys, then it'll have kik-i girls. How many ways are there to choose ii boys from nn boys? (ni)n\choose i.

How many ways are there to choose kik-i girls from mm girls? (mki)m\choose{k-i}.

So for a fixed ii, there are (ni)(mki){n\choose i}{m\choose {k-i}} ways to form a team of kk students.

Since ii could be any number from 00 to kk, we add the number ways to form the team for different values of ii.

So, our final count is,

i=0k(ni)(mki)\displaystyle \sum_{i=0}^k {n\choose i}{m\choose{k-i}}

Since answers 1 and 2 are counting the same thing, they must be equal.


Mursalin Habib - 7 years ago

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Jianzhi Wang - 7 years ago

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yeah, that is the actual counting in 2 ways.

Abhishek Bakshi - 6 years, 12 months ago

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The last identity is known as Vandermonde's Identity.

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Deriving Vandermonde's identity is very simple . ( I am going to tell just the procedure ) First what you need to do is just right binomial expansion of (1+x)n(1+x)^{ n } . again rewrite the binomial expansion of(x+1)m(x+1)^{ m } . Note that you should expand (x+1)m(x+1)^{ m } not (1+x)m(1+x)^{ m } . now multiplying these two expansions . we get the above summation which is equal to the one of the binomial coefficient in the expansion of (x+1)m+n(x+1)^{ m+n } Hence prooved

Anish Kelkar - 6 years, 12 months ago

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That's funny that you used LaTeX\LaTeX on the whole thing. Yes, I will surely work on this cool identity. :D

Finn Hulse - 7 years ago

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Yay thanks :)

Victor Loh - 7 years ago

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How do I prove it in a non-combinatorial way?

Victor Loh - 7 years ago

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Consider the coeff of xkx^k of both sides in (1+x)n(1+x)m=(1+x)n+m(1+x)^n (1+x)^m=(1+x)^{n+m}

Abhishek Sinha - 7 years ago

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Thank you.

Victor Loh - 7 years ago

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