Mr.John has x children by his wife and Mrs.Sonia has (X+1) children by her husband. They got married and have been blessed with some children. Now , the whole family has 10childern. Assuming that two children of the same parents do not fight, find the maximum number of fights that can take place among children?
A.33 B.22 C.111 D.None of these.

• Mr.John has x children while Mrs.Sonia has (x+1) children. After they got married, totally they have 10 children. But it is not given how many children do they have after they got married. Let they be 'y'. Then, we get,
  x+(x+1)+y=10 i.e. 2x+y=9. Here, if y can take only odd values, otherwise, x will not be a positive integer for any even value of y.
  Then there are following cases:
  1) for y=1, x=4 and x+1=5,
  2) for y=3, x=3 and x+1=4
  3) for y=5, x=2 and x+1=3
  4) for y=7, x=1 and x+1=2
  5) for y=9, x=0 and x+1=1.
  As the children of same parents cannot quarrel, only y can quarrel with x and (x+1) and x can quarrel with (x+1). Then from above five cases, we have to consider the case which will give maximum value. It is the the 2nd case. That is, the product xy+y(x+1)+x(x+1) will be maximum for y=3, x=3 and (x+1)=4, that is 33.
  Hence 33 is the answer.
• 12 years agoHelpfull: Yes(38) No(0)
• total children = (x+x+1)= 2x+1
  now 2x+1!=10 because some new Children (more than 1)born.
  thus 2x+1=7/11 but 11 is out of condition.
  thus
  3,3,4 are children of two different family.
  however 2 children of same family cant fight.thus possibilities are:
  3c1.3c1+3c1.4c1+4c1.3c1=33
  
  tell if correct?
  
  
• 12 years agoHelpfull: Yes(13) No(8)
• plz send me to0 at..
  chandana.pandey@gmail.com
• 12 years agoHelpfull: Yes(3) No(2)
• Ans: 7c3 = 35
  
  here only max combinations possible
• 12 years agoHelpfull: Yes(2) No(15)
• @navneet i think ur wrt....:) i got the same ans yar :)
• 12 years agoHelpfull: Yes(0) No(6)
• plz explain me i can't understand it ..
• 12 years agoHelpfull: Yes(0) No(0)
• first combination of people 3,4,3 gives maximum nuber of fights because fist 3 each one have 7 option for fight therfore there is 3*7=21.now come to 2nd 4 children have each children have 6 option so total 6*4=24.now come to 3rd similar to first 3 children last 3 chidren have 21 option for fight .so total we have 21+24+21=66 options which is maximum.
  
  so option d is correct which is none of these
• 11 years agoHelpfull: Yes(0) No(0)