Zada has to distribute 15 choclates among 5 of her children sana,ada,amir,farhan. she has to make sure that sana get atleast 3 and atmost 6 choclates.in how many ways can this be done?

• there would be 4 cases
  1.A gets 3 chocolates then the other 12 can be distributed among other 4 children in (12-1)C(4-1)ways ,ie;165 ways
  2.A gets 4 chocolates then the other 11 can be distributed among other 4 children in (11-1)C(4-1)ways ,ie;120 ways
  3.A gets 5 chocolates then the other 10 can be distributed among other 4 children in (10-1)C(4-1)ways ,ie;84 ways
  4.A gets 6 chocolates then the other 9 can be distributed among other 4 children in (9-1)C(4-1)ways ,ie;56 ways
  
  adding all above cases,we get 165+120+84+56=425 ways.....
  
  hence the solution is 425 ways.
• 11 years agoHelpfull: Yes(42) No(5)
• In this we don't use the formula (n-1)c(r-1). Because this formula is distributing
  n identical things among r people/groups such that any person/group might get any number(except 0).
  But here the others children might get zeros no. of chocolate.
  So we use the formua (n+r-1)c(r-1). so we get the answer 888.
• 11 years agoHelpfull: Yes(12) No(6)
• 435 ways of distributing the chocolates
• 11 years agoHelpfull: Yes(10) No(12)
• you people are assuming that each child gets atleast 1 chocolate but it can be a case that a child get no chocolate..so in my opinion the answer would be 15c3+14c3+13c3+12c3
  what you guys say???
• 11 years agoHelpfull: Yes(9) No(4)
• sana gets 3,and then we give 1 each to remaining 4 so, 7 chocolates over we have 8 chocolates remaining.now we need to distribute the remaining 8 to the other four children.this can be done in 8+4-1C4-1 ways [case 1]. In line with this we also have cases 2,3,4.
  
  case 1 when sana gets 3 chocolates: ( 8+4-1)C(4-1) = 11C3 = 165
  
  case 2 when sana gets 4 chocolates 10C3 = 120
  
  case3 sana gets 5 chocolates 9C3 = 84
  
  case4 sana get 6 chocolates 8C3 = 56
  
  thus total ways are 165 + 120+ 84+56 = 425
• 10 years agoHelpfull: Yes(7) No(0)
• 11c3+10c3+9c3+8c3=425
• 11 years agoHelpfull: Yes(4) No(3)
• is it actually 5 children or 4 children ?
  
• 11 years agoHelpfull: Yes(4) No(0)
• there is a problem in this que
  it is not mention that each one get atleast one thing , here we can take zero object for any one or not , i think we can
• 11 years agoHelpfull: Yes(4) No(0)
• answer is 12-1c94-1)+11-1c(4-1)+10-1c(4-1)+9-1c(4-1)=425
  
• 11 years agoHelpfull: Yes(3) No(2)
• 15c3+14c3+13c3+12c3..this vl b the correct ans..this combination gives all possible ways
• 11 years agoHelpfull: Yes(3) No(1)
• lets satisfy sana by giving her 3 chocolate first and 1 chocolate each to remaining 4 children so total choc remaining will be 8... now this 8 chocolate can be distributed among 5 children in { 8+5-1)C(5-1) = 495 ways .. now this formula (n+r-1)C(r-1) means distributing n identical things among r thing such that these r can get any number of things including 0....
  which means in 495 sana may have got 0,1,2,3,4,5,6,7 or all 8 chocolates... but this is wrong as sana cannot have more than 6 chocolates so we have to eliminate the no of ways through which she got more than 6 chocolates..... that cases are when she got 8 or 7 or 6 or 5 or 4 chocolates (as she already have 3 chocolates so 3+3 is okay but 3+4i=7 is wrong)
  NOW
  1)she can get 8 more chocolate in only 1 case=1
  2)she can get 7 more chocolate in 4C1=4 (4C1 because in this case 1 choc is remaining to distribute among 4 childs.... n=4 r=1 so 4C1)
  3)she can get 6 more chocolate in 5C1=5 (because now now 2 are remaining among 4 childs.... 4+2-1C2-1)
  4)she can get 5 more chocolate in 6C2=15 (because now now 3 are remaining among 4 childs.... 4+3-1C3-1)
  5)she can get 4 more chocolate in 7C3=35 (because now now 4 are remaining among 4 childs.... 4+4-1C4-1)
  .
  .
  .
  total making 60 cases where she can get 6+ chocolates thus
  295-(1+4+5+15+35)=295-60=235 cases
  ANS:235 and it is the correct answers for sure.......
  
• 9 years agoHelpfull: Yes(2) No(7)
• book me to 435 hai awnser
  ?
  can someone please help me in this
• 7 years agoHelpfull: Yes(1) No(0)
• arun sharma que
  
• 11 years agoHelpfull: Yes(0) No(1)
• 12-1c94-1)+11-1c(4-1)+10-1c(4-1)+9-1c(4-1)=425
• 10 years agoHelpfull: Yes(0) No(1)
• 12c4+11c4+10c4+9c4
  
• 8 years agoHelpfull: Yes(0) No(0)