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#3**+10 **

Not a "proof"...but notice....

The rows of Pascal's triangle always sum to 2^{n} 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 = 2^{5} = 32

But notice also, that this is just (1 + 5 +10)*2 = (16)*2 = 2^{4 *} 2 = 2^{5}

And every row has this symmetry of 2^{n-1} * 2 = 2^{n} ....(note that row "0" = 2^{0-1} * 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 = 2^{2} = 4

Or row 6..... 1, 6, 15, 20, 15, 6, 1 = (1 + 6 + 15 + 10)*2 = 2^{5} * 2 = 2^{6}

CPhill Dec 25, 2014

#1**+10 **

The number of possible subsets that can be formed from "n" elements in a set is just 2^{n}

So 2^{5} = 32 possible subsets

CPhill Dec 25, 2014

#2**+10 **

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?**

Melody Dec 25, 2014

#3**+10 **

Best Answer

Not a "proof"...but notice....

The rows of Pascal's triangle always sum to 2^{n} 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 = 2^{5} = 32

But notice also, that this is just (1 + 5 +10)*2 = (16)*2 = 2^{4 *} 2 = 2^{5}

And every row has this symmetry of 2^{n-1} * 2 = 2^{n} ....(note that row "0" = 2^{0-1} * 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 = 2^{2} = 4

Or row 6..... 1, 6, 15, 20, 15, 6, 1 = (1 + 6 + 15 + 10)*2 = 2^{5} * 2 = 2^{6}

CPhill Dec 25, 2014