Question

c(n,1)+c(n,2)+....+c(n,n) kiske barabar he​

Answer 1

Answer:

ye hai kya mere ko to samajh hi nahi aa raha hai

Explanation:

Answer 2

Answer:

MARK AS BRAINLIAST ANSWER

Explanation:

There are two easy ways.

As Larry Pablo Storeling said, expand out (1+1)^n.

For a set of n element, there are 2^n subsets (as each element can be present or absent in any given subset). The number of subsets of size 0 is C(n,0), the number of size 1 is C(n,1), the number of subsets of size 2 is C(n,2), etc. So if you add them all up, you get all the subsets, i.e. 2^n.

The methods behind both of these proofs are quite extendable to other situations, so it’s worth keeping both in mind.