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`        JAMES HAS x CHILDREN WITH HIS FIRST WIFE. MARY HAS x+1 CHILDREN WITH HER FIRST HUSBAND .THEY MARY EACH OTHER AND HAVE CHILDREN OF THEIR OWN .THERE ARE TOTAL 24 CHILDRENS. ASSUMING THAT NO TWO CHILDREN  OF SAME PARENTS FIGHT WITH EACH  OTHER. HOW MANY TOTAL FIGHTS THERE WILL BE............. `
7 years ago

147 Points
```										Dear rahul
let james and mary have y children
x+(x+1) +y=24
y= 23-2x
so total number of fight =24C2
these fights also include firght among the children of same parents.
so children of james and his first wife can fight =xC2  ways

and children of Mary and her first husband  can fight =x+1C2  ways

children of james and mary can fight =yC2 = (23-2x)C2 ways
so total number of desired fight

24C2 -xC2 -x+1C2  -(23-2x)C2
simplify  -3x2 +45x+23
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7 years ago
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