Zada has to distribute 15 choclates among 5 of her children Sana,ada,jaya,amir and farhan. She has to make sure that sana gets at least 3 and ada at most 6 choclates. In how many ways can this be done?
A.495 B.77 C. 417 D. 435

• Actual Ans is : 425
  
  1st slot : 4 choco are distributed. remains 15-4=11
  case 1 when san get 3 chocolate: ( 8+4-1)C(4-1) = 11C3 = 165
  
  case 2 when sen get 4 chlt 10C3 = 120
  
  case3 sen get 5 chlt 9C3 = 84
  
  case4 sen get 6 chklt 8C3 = 56
  
  thus total ways are 165 + 120+ 84+56 = 425
• 9 years agoHelpfull: Yes(2) No(1)
• This qn is present in arun sharma --> Permutation & combination --> qn. 64 lod 1
• 9 years agoHelpfull: Yes(1) No(0)
• D)435
  check
• 9 years agoHelpfull: Yes(0) No(0)
• Solve it Santosh
• 9 years agoHelpfull: Yes(0) No(0)
• Does this qn came in ur elitmus exam ???
• 9 years agoHelpfull: Yes(0) No(0)
• ans 435
  check
• 9 years agoHelpfull: Yes(0) No(0)
• 495
  (3,1,11)=165
  (4,1,10)=120
  (5,1,9)=84
  (6,1,8)=56
  
• 9 years agoHelpfull: Yes(0) No(0)
• @avijitsharma... derz no sch ques in arun sharma quant apti.......can u plz explain it??
• 9 years agoHelpfull: Yes(0) No(0)
• vishaka garg -- yes..it is there .... on page 583...anyways...here is the solution:
  
  We will first give 3 chocolates to Sana and 1 each
  to the other 4. So we have already distributed -
  chocolates. Now we are left with 8 chocolates that
  have to be distributed among 5 people. So number
  of ways possible is 12!/8!4! -- 495. Out of these
  we have to eliminate cases in which Sana gets
  more than 6 chocolates. As we have already given
  her 3 chocolates that means we have to eliminate
  cases in which she gets 7 or 8 or 9 or l0 or
  11 chocolates out of 495 cases. She would get 11
  chocolates in one case, l0 chocolates in 4C1 cases,
  9 chocolates in 5C1, cases, 8 chocolates in 6C2 cases
  and 7 chocolates in 7C3, cases. So, total number of
  cases that need to be eliminated
  is 60. So the required
  number of ways is 495 - 60 =
  435
• 9 years agoHelpfull: Yes(0) No(0)