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Binomial Coefficients


We know that,

        (a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 +...+ nCn xn                ... (i)

Let us write equation (i) in particular form by putting a = 1. Now it becomes,

        (1 + x)n = nC0 + nC1 x + nC2 x2 +...+ nCr xr +...+ nCn xn                   ... (ii)

Coefficients attached with different powers of x are called Binomial Coefficients.

Now, again put a = 1 in the expression

(a + x)n = an + n an-1 x + (n(n-1))/2! an-2 x2 + (n(n-1)(n-2))/3! an-3 x3

+...+ (n(n-1)(n-2)....(n-r+1))/r! an-r xr +......+ xn

we get,

        (1+x)n = 1 + nx + (n(n-1))/2!  x2 +...+ (n(n-1)(n-2)......(n-r+1))/r!  xr +...+ xn            ...(iii)

        If we compare coefficient of xr in expansions (ii) and (iii), we have,

                nCr = (n(n-1)(n-2)......(n-r+1))/r! = n!/r!(n-r)!

What does it mean when we compare two expansions?

        Since expansion (ii) is valid for any value of x, we can replace x by 1/x , (x ≠ 0) in it, we get,

        (1 + (1/x))nC0 + nC1 1/x + nC2  1/x2 +...+ nCr  1/xr +...+ nCn  1/xn

        Multiplying both sides by xn, it becomes,

        (1+x)n = nC0 xn + nC1 xn-1 +...+ nCr xn-r +...+ nCn                        ... (iv)

        Compare expansion (ii) and (iv). It will give us, that

       nCr = nCn-r, this is a beautiful result, which says that in the expansion of the equations of the like of (ii) the rth coefficient from the beginning is equal to the rth coefficient from the end.

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