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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.............

rahul khurana

16 Years agoGrade

1 Answer

Badiuddin askIITians.ismu Expert

16 Years ago

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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