Seven different objects must be decided among 3 persons. In how many ways this can be done if at least one of them gets exactly one object???

• n-1c r-1 =6c2=15
• 9 years agoHelpfull: Yes(15) No(7)
• Division of m+n+p objects into three groups is given by (m+n+p)/(m!×n!×p!)
  so, according to question at least one gets only one object this can be done in (1,1,5),(1,2,4),(1,3,3) ways.
  
  required number of ways= (7!/(1!*1!*5!))*(1/2!) + (7!/(1!*3!*3!))*(1/2!) + (7!/(1!*2!*4!)) = 196
• 9 years agoHelpfull: Yes(11) No(3)
• (1) suppose only 1 person gets 1 object.
  so, the other two get either (0,6) or (2,4) or (3,3) objects.
  so, no. of ways of partitioning the remaining 6 objects into each such group = 1, 15, 20 respectively
  req no. of ways this can happen = 7*3*(1+15+20)*2
  
  (2) now suppose only 2 persons get 1 object each
  so, the third must get remaining 5
  req no. of ways this is possible = 3*21
  
  add up the nos.
• 9 years agoHelpfull: Yes(2) No(2)
• (7!/(1!*1!*5!))*(1/2!) + (7!/(1!*3!*3!))*(1/2!) + (7!/(1!*2!*4!)) = 196
• 9 years agoHelpfull: Yes(2) No(0)
• Answer must be like in that way...
  They have already said different object so first take a,b,c three people ,a get 1 object in 7 way .and remaining 6 obj can be divided as 6c2 way .as they diffrnt ..so thats 7x6c2=105 ways..and this can be happened three way.. so ans in 315 .
• 9 years agoHelpfull: Yes(1) No(2)
• 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
  so the answer is 196
• 9 years agoHelpfull: Yes(1) No(1)
• formula for this is( m+n+p)!/m!n!p!
  so 7!/3!=140 ways
  
• 9 years agoHelpfull: Yes(0) No(4)
• 7 different objects must be decided among 3 persons.
  so,no of ways to get by exactly one person,exactly one object = 7C3 -1
• 9 years agoHelpfull: Yes(0) No(1)
• 7c3=7*6*5*4*3/1*2*3
  =420
• 9 years agoHelpfull: Yes(0) No(1)
• 6 items can be divided into 6c2=15ways
  and anyone gets one item
  so total no ways =15*3=45
  
• 9 years agoHelpfull: Yes(0) No(1)
• 7p3 is the right answer
• 9 years agoHelpfull: Yes(0) No(2)
• 7c3 - 3 - 0 = 7c3 - 3
  7c3 = 7 different objects decided among 3 persons.
  3 = ways in which 1 object can be given to 2 men.
  0 = ways in which 1 object can be given to 3 men
  So the remaining are the ways in only 1 man get 1 object
• 9 years agoHelpfull: Yes(0) No(0)
• (1) suppose only 1 person gets 1 object.
  so, the other two get either (0,6) or (2,4) or (3,3) objects.
  so, no. of ways of partitioning the remaining 6 objects into each such group = 1, 15, 20 respectively
  req no. of ways this can happen = 7*3*(1+15+20)*2
  
  (2) now suppose only 2 persons get 1 object each
  so, the third must get remaining 5
  req no. of ways this is possible = 3*21
  
• 9 years agoHelpfull: Yes(0) No(1)
• lets take it like this _ _ _ .3 people have to get from 7 objects. In question it has stated that one has to get exactly one object so 1st person gets one then remaining 2 person has to get from 6 objects
  
  possibilities
  2nd person 6
  3rd person 5
  so 1*60*5 = 30
  
  Ans:30
• 9 years agoHelpfull: Yes(0) No(0)
• The objects can be divided into three groups that are (1,1,5) or (2,4,1) or (3,3,1) as atlas one person should get 1 object.
  so number of ways can be [(7C1*6C1*5C5)/2!]+(7C2*5C4*1C1)+[(7C3*4C3*1C1)/2!]=196
• 9 years agoHelpfull: Yes(0) No(0)
• case1: * * *
  1 object for one person and remaining 6 objects for 2 persons=6c2=15
  three cases are possible so 3*15=45
  case2: for 2 persons each one object then fr one person 5 objects
  3 cases are possible so 3*5=15
  hence 15+45=60
• 9 years agoHelpfull: Yes(0) No(0)
• there 7 objects
  so 1 give to each
  4 remains
  as per product and sum rule 3*(4+3+2+1)=3*10=30
  ans:30
• 9 years agoHelpfull: Yes(0) No(0)
• there are 3 possible dist. scenario. (1,1,5),(1,2,4),(1,3,3)
  (1,1,5)
  since seven objects are different, not identical, thus no of ways of dist.ing them = (7C1+6C1+5C5)*(3!/2!)=126
  similarly,
  for(1,3,3)
  (7C1*6C3*3C3)*(3!/2!)=420
  for(1,2,4)
  (7C1*6C2*4C4)*(3!)=630
  hence total ways=(630+420+126)=1176
  
• 9 years agoHelpfull: Yes(0) No(0)
• Answer is 15
  because
  x+y+z=7
  question says at least one
  (1,1,5)=this can be done in 3!/2!=3 or(+)
  (2,2,3)=this can be done in 3!/2!=3 or(+)
  (1,3,3)=this can be done in 3!/2!=3 or(+)
  (1,2,4)=this can be done in 3! = 6
  so finally sum is 15
  :)
• 9 years agoHelpfull: Yes(0) No(0)
• 1 2 4 3! =6
  1 3 3 3!/2=3
  1 1 5 3!/2=3
  6+3+3=12
  
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 18 ..because atliest one of them get one object!!!
• 8 years agoHelpfull: Yes(0) No(0)