Question

The sum of all possible combinations of n distinct things is 2 n. C0 + nC1 + nC2 + . . . + nC n =​

Answer 1

Answer:

In the binomial expansion

(a+b)^n = nC0a^n. b^0 + nC1 a^(n-1)b^1 + nC2a^(n-2)b^2 + etc

If you now let a=b=1

(1+1)^n =2^n = nC0 + nC1 +nC2 +. etc .

In fact what we have here is the number of subsets of size r taken from a set of n objects.

nC0 = 1 = number of sets containing zero elements

nC1 = n = " ". " " 1. "

nC2 = (n)(n-1)/2 ". ". ". ". 2 ". etc

Total = nC0 + nC1+ nC2 +. = 2^n =total number of subsets in a set of n elements.

Again if a=1 and b=-1

(1-1)^n =0 = nC0 - nC1 + nC2 - nC3

nC0 +n C2 + nC4 + = nC1 + nC3 + nC3 +.

= (1/2) sigma nCr = (1/2) .2^n = 2^(n-1)