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question asked: 11 Oct, 07:46
question was seen: 66 times
last updated: 11 Oct, 22:44
there are only 2 cases possible
(x,x,x,a+b+c=x)
(x,x,a+b=x,c+d=x)
for 1st:
so lets just go through all possible x,
find number of ways such that a+b+c=x,lets say it is val
so ans of first type ans1=((no.of.x)C3)*val
for 2nd:
lets just go through all possible x,
find number of ways of ways of choosing 4 numbers a,b,c,d such that ((a+b)=x and (c+d)=x) or ((a+c)=x and (b+d)=x) or ((a+d)=x and (b+c)=x)
let this be val3
so ans of second type ans2=((no.of.x)C2)*val2
res=ans1+ans2
try thinking of some dp for val,val3.....