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Meaning of the binominal coefficient
By Yatir Halevi on Wednesday, April 17, 2002 - 08:45 pm:

If we take the binominal coefficient C(n,k)=n!/(k!(n-k)!) in it's combinatorical meaning, how do we explain the reason it appears in the expansion of (a+b)n?


Thanks,
Yatir

By Andre Rzym on Friday, April 19, 2002 - 07:26 pm:

Yatir,

I believe that the appearance of C(n,r) coefficients in the expansion of (a+b)n can be explained directly in terms of the combinatorial meaning of C(n,r) (i.e. the number of ways of choosing r items from a collection of n items without repetition and without regard to the order in which the r items are selected).

Consider

(a+b)n=(a+b)(a+b)...(a+b) n times

Suppose we expand the right hand side , but do not gather together terms of equal powers. Then we get:

S(a or b from first bracket)*(a or b from second bracket)*...*(a or b from n'th bracket)

The sum is over all the different "a or b's". Given that there are n brackets, 2 options per bracket, the sum gives rise to 2n terms.

We observe that if two of the terms in this series have the same number of a's, they must also have the same number of b's and can therefore be combined if we wish.

Here's the crux of the matter: How many terms in the sum have exactly r a's in them, for that is the coefficient of arbn-r. This is a combinatorial question:"From the n parentheses, in how many ways can we choose r of them (to be a's) without regard to the order in which we selected the parentheses?". The answer, of course, is C(n,r).

Yatir, are you convinced?

Andre

By Yatir Halevi on Sunday, April 21, 2002 - 08:53 pm:

Very much so...

Thanks, Andre!

Yatir