Question

is there a more discrete math way to answer this?
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Answer 1

given the sets x,y,z

let x= {1,2,3,4}
let y = {2,3,5,6}
let z = {3,4,5,7}

if we add them up with no regard for thier intersections, we are 'duplicating' the data.

(1+2+3+4)+(2+3+5+6)+(3+4+5+7)

we need to adjust it to get an accurate account, subtract the intersections xy, xz, yz

(1+2+3+4)+(2+3+5+6)+(3+4+5+7)
-(2+3) -(3+4) -(3+5)

what we have done is; 2+2-2, 4+4-4, 5+5-5, and 3+3+3 -3-3-3

the 2,4,5 have been taken care of so that they are counted once ... but we have completly eliminated the 3s ...

add back in the intersection xyz ....

(1+2+3+4)+(2+3+5+6)+(3+4+5+7)
-(2+3) -(3+4) -(3+5)
+(3)

now everything is accounted for and there is no duplication of the data.