in an exam the maximum mark for each of 3 subject is 50 and the maximum mark for the forth subject is 100 .find the number of ways in which a student can score 60% marks in aggregate.

• 3C3 or (1C1 and 3C1)
  =3C3+1C1*3C1
  =1+3
  =4 ways
• 9 years agoHelpfull: Yes(4) No(4)
• 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(2) No(6)
• LET THE SCORE BE
  a/50
  b/50
  c/50
  d/100
  max marks=250
  60% 0f 250=150
  ie. a+b+c+d =150
  ie 153C3 is the answer
• 9 years agoHelpfull: Yes(1) No(17)
• please provide correct solution
• 5 years agoHelpfull: Yes(0) No(0)
• Correct option is
  B
  110551
  The candidate must score 150 marks.
  ∴ Required number
  =coefficient of x
  150
  in (1+x+...+x
  50
  )
  3
  (1+x+...+x
  100
  )
  = coefficient of x
  150
  in (
  1−x
  1−x
  51
  
  ​
  )
  3
  
  1−x
  1−x
  101
  
  ​
  
  =coefficient of x
  150
  in (1−x
  51
  )
  3
  (1−x
  101
  )(1−x)
  −4
  
  =coefficient of x
  150
  in (1−3x
  51
  +3x
  102
  −x
  153
  )(1−x
  101
  )(1−x)
  −4
  
  [leaving terms containing powers of x greater than 150]
  =coefficient of x
  150
  in (1−x)
  −4
  −3.
  =coefficient of x
  99
  in (1−x)
  −4
  +3coefficient of x
  48
  in (1−x)
  −4
  coefficient of x
  49
  in(1−x)
  −4
  
  =
  153
  C
  150
  ​
  −3.
  102
  ​
  C
  99
  ​
  +3.
  51
  C
  48
  ​
  −
  52
  C
  49
  ​
  
  =
  6
  153.152.151
  ​
  −3.
  6
  102.101.100
  ​
  +3.
  6
  51.50.49
  ​
  −
  6
  52.51.50
  ​
  
  =110551
• 3 years agoHelpfull: Yes(0) No(0)