Not a "proof"...but notice....
The rows of Pascal's triangle always sum to 2n where n is an integer ≥ 0
And any row "n" will also equal the sum of C(n,m) where "m" ranges from 0 to n
So, for the fifth row, we have 1, 5, 10, 10, 5, 1 = 32 = 25 = 32
But notice also, that this is just (1 + 5 +10)*2 = (16)*2 = 24 * 2 = 25
And every row has this symmetry of 2n-1 * 2 = 2n ....(note that row "0" = 20-1 * 2 = 2-1 * 2 = 1/2 * 2 = 1)
The even rows can be "split".....for instance.....row 2 is 1, 2, 1 = 1 + (1 +1) + 1 = (1 + 1)*2 = (2)1*2 = 22 = 4
Or row 6..... 1, 6, 15, 20, 15, 6, 1 = (1 + 6 + 15 + 10)*2 = 25 * 2 = 26
The number of possible subsets that can be formed from "n" elements in a set is just 2n
So 25 = 32 possible subsets
Mmm
zero elements 1
one element 5
two elements 5C2=10
3 elements 10
4 elements 5
5 elements 1
TOTAL = 32
Mmm same answer as CPhill. Now the question is "why are these the same"
Who want to provide a proof that these methods will always result in the same answer?
Not a "proof"...but notice....
The rows of Pascal's triangle always sum to 2n where n is an integer ≥ 0
And any row "n" will also equal the sum of C(n,m) where "m" ranges from 0 to n
So, for the fifth row, we have 1, 5, 10, 10, 5, 1 = 32 = 25 = 32
But notice also, that this is just (1 + 5 +10)*2 = (16)*2 = 24 * 2 = 25
And every row has this symmetry of 2n-1 * 2 = 2n ....(note that row "0" = 20-1 * 2 = 2-1 * 2 = 1/2 * 2 = 1)
The even rows can be "split".....for instance.....row 2 is 1, 2, 1 = 1 + (1 +1) + 1 = (1 + 1)*2 = (2)1*2 = 22 = 4
Or row 6..... 1, 6, 15, 20, 15, 6, 1 = (1 + 6 + 15 + 10)*2 = 25 * 2 = 26