How many ways can 7 different objects be distributed among 3 people such that at least one of them gets exactly one object?

• 7-1C3-1=6C2=15
• 9 years agoHelpfull: Yes(30) No(15)
• Ans : 12
  bcz we'll not consider the case (2,2,3) as the question says that there should be exactly one onject gets at least one. So,
  (1,1,5) = 3!/2! = 3
  (1,3,3) = 3!/2! = 3
  (1,2,4) = 3! = 6
  3 + 3 + 6 = 12
• 9 years agoHelpfull: Yes(17) No(12)
• 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!x2! + (7)!/ 1!×2!×4! + (7)!/ 1!×1!×5!x2! = 70 + 105 + 21 =
  
  196
• 9 years agoHelpfull: Yes(6) No(5)
• ans 381
  (3*2^7)-3=381
• 9 years agoHelpfull: Yes(6) No(8)
• 7-1c3-1=6c2=15
• 9 years agoHelpfull: Yes(5) No(0)
• 6!/2!*4!=15
• 9 years agoHelpfull: Yes(4) No(5)
• 2^7*3-3=381
  
• 9 years agoHelpfull: Yes(4) No(5)
• There will be double counting if u do 2^6 to split the remaining 6 having alr done 7x3 to give one object to one of the individuals. Eg. Individuals A,B,C, assume u chose A to be the one to have exactly one u can have 1,5,1 as one outcome. If u chose C to have exactly one in the first place and also do 2^6, u can also get 1,5,1. Thus, there is double counting. Correct ans is 1218
• 8 years agoHelpfull: Yes(4) No(0)
• 1218 it is :)
• 9 years agoHelpfull: Yes(2) No(4)
• ans-12.distributed as
  so,
  115,151,511
  124,241,412,421,214,142
  133,331,313
  toatal distribusion=12
• 9 years agoHelpfull: Yes(2) No(1)
• a person to get exactly one object can be chosen in 3C1 ways ,
  then objects left will be 6, they have to be distributed among two persons,
  so an object can be distributed in 2 ways, so 2*2*2*2*2*2 i.e. 2^6,
  therefore, as per this, ans will be 2^6+3C1 = 192. please let me know the correct answer.
• 9 years agoHelpfull: Yes(2) No(0)
• I am 100% sure that answer will be = 1218 .
  Details EXplan :- case (1): if only one gets one object
  The remaining can be distributed as: (6,0),(4,2),(3,3).
  (7C1*6C6*3!+7C1*6C4*3!+7C1*6C3*3!/2!) = 42 + 630 + 420 = 1092 ;
  Case (2): If 2 people get 1 object each:
  7C1*6C1*5C5*3!/2! = 126;
  Thus, A Total of 1218..
  hope it is easy to understand
• 8 years agoHelpfull: Yes(2) No(0)
• Ans : 12
  (1,1,5) = 3!/2! = 3
  (1,3,3) = 3!/2! = 3
  (1,2,4) = 3! = 6
  3 + 3 + 6 = 12
• 9 years agoHelpfull: Yes(1) No(3)
• there must be atleast 1 in each
  (ie)3,2,2=3!/2!=3 should not be applicable.
  1,1,5=3!/2!=3
  4,2,1=3!=6
  1,3,3=3!/2!=3
  
  Total ans=3+6+3=12
  
• 9 years agoHelpfull: Yes(1) No(3)
• ans:18
  bcz 1,0,6=3!
  1,1,5=3
  1,2,4=3!
  1,3,3=3
• 9 years agoHelpfull: Yes(1) No(5)
• 7c1.6c6.3!+7c1.6c5.3!/2!+7c1.6c4.3!+7c1.6c3.6c3.3!/2!=1218
• 9 years agoHelpfull: Yes(1) No(4)
• atlest one of them so,(let A,B,C=3)
  and 7c1+7c2+7c3+7c4+7c5+7c6=2^7-2=126
  therefore ,126*3=378 ways
• 9 years agoHelpfull: Yes(1) No(1)
• i think surabhi's answer is correct
• 9 years agoHelpfull: Yes(1) No(0)
• we can distribute 7 objects among 3 people
  1.(1,1,5)
  2.(1,2,4)
  3.(1,3,3)
  7C1*6C1*5C5 + 7C1*6C2*4C4 +7C1*6C3*3C3 = 287
• 8 years agoHelpfull: Yes(1) No(0)
• Friends, This will be like this:
  7 objects are there and 3 has to get at least one. So give one to each.Now we have 4 more to be distributed among 3. So follow
  m+n-1cn-1(Arun Sharma given formula.)
  4+3-1c3-1=6c2
  =6*5/2=15
• 8 years agoHelpfull: Yes(1) No(0)
• 1,1,1
  1,1,2
  1,1,3
  1,1,4
  1,1,5
  for one place it's 5 and for three places it will be 15
  
• 9 years agoHelpfull: Yes(0) No(6)
• (1,1,5) (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
• 9 years agoHelpfull: Yes(0) No(1)
• (1,1,5) (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
  (1,1,5)
• 9 years agoHelpfull: Yes(0) No(0)
• (1,1,1)---> 1 way
  (1,1,5)---> 3 ways
  (1,4,2)---> 6 ways
  (1,3,3)---> 3 ways
  total 13 ways
• 9 years agoHelpfull: Yes(0) No(0)
• I think it would be 63..
  Bcz one gets exactly one..let us assume A,B,C are there.. so that the way of getting object to atleat one is 21.. each 3 person gets 7 differnt object simultaneously..
  Next consider the rest two.. they will get the 7 objects in 6 different ways..
  So the sum is= (6*7)+(3*7)=63
• 9 years agoHelpfull: Yes(0) No(0)
• 3*(2^6-1)=189
  no of ways of selecting the one who gets exactly one object is= 3c1
  thus remaining 2 people will get from the remaining 6 objects= 2^6-1
  thus total = 3c1*(2^6-1)
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 210 because 1 peron gets 7 out 7 objects and 2 person gets 6 next 5
• 9 years agoHelpfull: Yes(0) No(0)
• @satish.....plz tell the correct answer for this
• 8 years agoHelpfull: Yes(0) No(0)
• http://gmatclub.com/forum/1-7-different-objects-must-be-divided-among-3-people-in-how-7409.html
  check out the ans its 2187
• 8 years agoHelpfull: Yes(0) No(1)