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

• if 2 of them don't get any, all object goes to 3rd person => 3 ways(A/B/C)
  
  if 1 of them don't get any, 7 objects goes to rest 2 persons (AB,BC,CA)
  this can be in 3 ways
  (1,6)=> 7c1
  (2,5)=> 7c2
  (3,4)=> 7c3
  (4,3)=> 7c4
  (5,2)=> 7c5
  (6,1)=> 7c6
  
  (7c1+7c2+7c3+7c4+7c5+7c6)= 2^7-2 = 126
  total ways = 3*126 = 378
  
  Total = 3 + 378 = 381
• 9 years agoHelpfull: Yes(46) No(3)
• Case 1: Two guys don't get anything. All items go to 1 person. Hence, 3 ways.
  
  Case 2: 1 guy doesn't get anything. Division of 7 items among 2 guys. Lets say they are called A & B.
  
  a) A gets 6. B gets 1. 7 ways.
  b) A gets 5. B gets 2. 21 ways.
  c) A gets 4. B gets 3. 35 ways.
  d) A gets 3. B gets 4. 35 ways.
  e) A gets 2. B gets 5. 21 ways.
  f) A gets 1. B gets 6. 7 ways.
  
  Total = 126 ways to divide 7 items between A & B.
  Total number of ways for case 2 = 126*3 = 378 (Either A, or B, or C don't get anything).
  
  Answer = Case 1 + Case 2 = 3 + 378 = 381
  
• 9 years agoHelpfull: Yes(12) No(1)
• IT CAN BE ALSO DONE IN THS WAY
  first case - 7,0,0 / 0,7,0 / 0,0,7 these are three ways
  second case -we have to keep only one zero i.e. to divide 7 objects to 2 of them which is 7p2 and we have to choose 2 out of three which is 3c2 so answer is
  3*7p2*3c2=381
• 9 years agoHelpfull: Yes(4) No(0)
• The possible cases in which either one or two of them do not get any object is:
  x$$,$$x,$x$,xx$,$xx,x$x i.e 6 different cases
  x means no object received
  $ means at least one object received
  Now for xx$,$xx,x$x cases where only 1 person receives the object, there is only 1 way of dividing the objects i.e. we have to give all the 7 objects to 1 person only. So, for these three cases we can divide the objects in 3 ways.
  Now for x$$,$$x,$x$ cases where only 2 persons receive the objects, there can be different possibilities to divide the 7 objects amongst them. It can be done in these manner:
  (1,6)(6,1)i.e 6*1+1*6 = 12 ways
  (2,5)(5,2)i.e 2*5+5*2 = 10 ways
  (3,4)(4,3)i.e 3*4+4*3 = 24 ways
  So, total such ways = 12+10+24 = 46
  Thus, 3+46 = 49 ways (Ans)
• 9 years agoHelpfull: Yes(3) No(5)
• abe Rakesh tere ko koi or suject ata hai ki bs maths me hi tees maar khan hai......
• 9 years agoHelpfull: Yes(2) No(3)
• Rakesh.. I don't understand how u write (1,6) => 7c1 ??
• 9 years agoHelpfull: Yes(1) No(1)
• @GOKUL thanks bhai
• 9 years agoHelpfull: Yes(0) No(0)
• 28 for the condition wen 1 of them didnt get is 7, for the condition 2 of them didnt get is 21
• 9 years agoHelpfull: Yes(0) No(1)