john has x children and mary has x+1children already.after marrying each other they have y children find the total no of fights among them if chidren of same parents do not fight with each other.total no of children are 24.
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10
total nmber of children = x + x+1 + y = 2x + y + 1 = 24
So y = 23 - 2x
group x fight with x+1 group : x (x+1)
group x+1 fights with y group = (x+1) y
group y fights with x group = x y
total = x² + x + 2 x y + y
= x² + x + 2x (23 -2x) + 23 - 2x
= - 3 x² + 45 x + 23
So y = 23 - 2x
group x fight with x+1 group : x (x+1)
group x+1 fights with y group = (x+1) y
group y fights with x group = x y
total = x² + x + 2 x y + y
= x² + x + 2x (23 -2x) + 23 - 2x
= - 3 x² + 45 x + 23
Answered by
6
total nmber of children = x + x+1 + y = 2x + y + 1 = 24
so 2x+y=24-1
So y = 23 - 2x
group x fight with x+1 group : x (x+1)
group x+1 fights with y group = (x+1) y
group y fights with x group = x y
total = x² + x + 2 x y + y
= x² + x + 2x (23 -2x) + 23 - 2x
= - 3 x² + 45 x + 23
so 2x+y=24-1
So y = 23 - 2x
group x fight with x+1 group : x (x+1)
group x+1 fights with y group = (x+1) y
group y fights with x group = x y
total = x² + x + 2 x y + y
= x² + x + 2x (23 -2x) + 23 - 2x
= - 3 x² + 45 x + 23
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