We know that,

        (a + x)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1 x + ⁿC₂ a^n-2 x² +...+ ⁿC_n 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 +...+ ⁿC_r x^r +...+ ⁿC_n 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 a^n-1 x + (n(n-1))/2! a^n-2 x² + (n(n-1)(n-2))/3! a^n-3 x³

+...+ (n(n-1)(n-2)....(n-r+1))/r! a^n-r x^r +......+ xⁿ

we get,

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

        If we compare coefficient of x^r in expansions (ii) and (iii), we have,

                ⁿC_r = (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))ⁿ  =  ⁿC₀ + ⁿC₁ 1/x + ⁿC₂  1/x² +...+ ⁿC_r  1/x^r +...+ ⁿC_n  1/xⁿ

        Multiplying both sides by xⁿ, it becomes,

        (1+x)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 +...+ ⁿC_r x^n-r +...+ ⁿC_n                        ... (iv)

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

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