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.

2
by raghunathreddy

2014-08-09T11:03:53+05:30

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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

2014-08-09T12:32:01+05:30
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