+0

# find the number of subsets {-12,0,12,13,14}

0
583
4

find the number of subsets {-12,0,12,13,14}

Dec 25, 2014

#3
+101149
+10

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

Dec 25, 2014

#1
+101149
+10

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

So 25  = 32 possible subsets

Dec 25, 2014
#2
+101741
+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?

Dec 25, 2014
#3
+101149
+10

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

CPhill Dec 25, 2014
#4
+101741
+5

That is really neat Chris.  Thank you :)

Dec 26, 2014