In how many can 7 different objects be divided among 3 persons so that either one or two of them do not get any object?

• if one person is given no object then the number of ways of distributing objects among rest of the two would be:
  2*(7C1 + 7C2 + 7C3)
  which is 378
  and the number of ways only one person gets all object is 3
  so total ways are 378+3=381
  
• 9 years agoHelpfull: Yes(14) No(3)
• 3c1*((2^7)-2)=378
• 9 years agoHelpfull: Yes(2) No(1)
• Whn two do not get nythn::
  A B C
  7 :1 way
  7 :1 way
  7 :1 way
  total 3 ways
  when 1 do not get nythn::
  6 1 :1 way
  1 6 :"
  6 1 :"
  1 6 :"
  6 1 :"
  1 6 :"
  so total for 6&1=6 ways
  similarly for 5&2=6 ways
  similarly for 4&3=6 ways
  so total=3+6+6+6=21 ways
• 9 years agoHelpfull: Yes(1) No(0)
• 381 is correct answer
• 9 years agoHelpfull: Yes(1) No(0)
• if one person gets no object then remaining two will get objects in following ways :
  3C2*(2^7-2)
  if two persons gets no object then remaining one persons gets all objects in ways:
  3C1*1
  So, total no of ways = 3C2*(2^7-2) + 3C1*1 = 381
• 8 years agoHelpfull: Yes(1) No(0)
• num of ways so that one person get nothing= 3c1*((2^7)-2))=368
  num of ways so that 2 person get nothing= 3c2*1=3
  so total num of ways=368+3=371
• 9 years agoHelpfull: Yes(0) No(2)
• x1+x2...+x7=3
  so for (n+r-1)c(r-1)=3+7-1c6=9c6=168
• 9 years agoHelpfull: Yes(0) No(1)
• ans
  n+r-1Cr-1 (n=7 , r = 3) = 36 ?
• 9 years agoHelpfull: Yes(0) No(2)
• can any one tell me why following process is not right??? i am not able to understand....
  
  (6c1*5c1*4c1*3c1)(14c1)/18c5=10/17
• 9 years agoHelpfull: Yes(0) No(0)
• Case 1: when one of them do not get any object, then the objects will be divided among 2 persons.
  
  No. of ways to select 2 persons = 3C2
  
  No. of ways to divide 7 objects among 2 persons = 27
  
  So required no. of ways= 3C2(27 - 2) = 378 [when none of them get any object]
  Case 2: when objects are given to only one person.
  
  Required. no. of ways= 3C1 = 3.
  
  So, total no. of ways = 378+3 = 381
  
  Hence 381is the correct answer.
• 8 years agoHelpfull: Yes(0) No(0)
• Case 1: when one of them do not get any object, then the objects will be divided among 2 persons.
  
  No. of ways to select 2 persons = 3C2
  
  No. of ways to divide 7 objects among 2 persons = 2^7
  
  So required no. of ways= 3C2(2^7 - 2) = 378 [when none of them get any object]
  Case 2: when objects are given to only one person.
  
  Required. no. of ways= 3C1 = 3.
  
  So, total no. of ways = 378+3 = 381
  
  Hence 381is the correct answer.
• 8 years agoHelpfull: Yes(0) No(0)
• 
  . Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object.
  
  Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p!
  But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
  So The number of ways are (7)!1!×3!×3!×12! + (7)!1!×2!×4! + (7)!1!×1!×5!×12! = 70 + 105 + 21 = 196
• 8 years agoHelpfull: Yes(0) No(0)