In an examination the maximum marks for each of the three papers is 50 each. The maximum marks for the fourth paper is 100. Find the number of ways with which a student can score 60% marks in aggregate.

1) 330850
2) 233551
3) 110551
4) 220800
5) none of these

• We need to distribute 60% marks ie. 150marks in 4 sections namely A,B,C,D.
  A+B+C+D=150
  
  Let 'a' be the marks that he loses in section A.. ie. (Marks obtained)+(Marks lost)=50 in section A
  Implies.. a+A=50
  a=50-A
  Similarly, b=50-B
  c=50-C
  d=100-D
  Now, A+B+C+D=150
  »(50-a)+(50-b)+(50-c)+(100-d)=150
  »a+b+c+d=100
  
  We need to distribute 100identical marks among 4 variables..
  (100+4-1)C(4-1)= 103!/100! 3!
  
  But this figure involves the cases when 'a' attains values like 51,52,53...up to 100.
  But this is not possible as section A can have maximum 50 marks.
  
  So, subtracting cases in which a=51,52,53....100
  
  Let a=51+ã, where ã can be any value starting from zero. So that we will get a=51+0 or a=51+1 or a=51+2... And so on.. All the values of 'a' will be covered.
  
  We know that, a+b+c+d=100
  »(51+ã)+b+c+d=100
  ȋ+b+c+d=49
  
  Distribution of 49 among 4 variables= (49+4-1)C(4-1)= 52!/49! 3!
  
  Here we have considered 'a' to be greater than 50.. Same goes when 'b' & 'c' will be greater than 50.
  Therefore multiplying above by 3.
  = 3x (52!/49! 3!)
  
  Final ans.=103!/100! 3! - 3x(52!/49!3!)
  =176851- 66300 = 110551
• 9 years agoHelpfull: Yes(11) No(0)
• In 1st three papers he can get marks 0,1,2,.......,50 and in fourth paper he can get 0,1,2,...........,100. In order to get 60% aggregate he needs 150 marks out of 250 total.
  
  Now, see carefully...
  
  if he get x1, x2, x3 and x4 marks in four papers then, the no. of ways to get 150 marks is same as the no. of integral soln of this eqn:
  x1 + x2 + x3 + x4 = 150
  with constraints, 0585276 - 515100 - 22100 + 62475
  =>110551
  
• 10 years agoHelpfull: Yes(2) No(8)