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.

• If only one get 1 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
  If 2 people get 1 object each:
  =7c1*6c1*5c583!/2!
  =126
  Total= 1092+126 = 1218
• 10 years agoHelpfull: Yes(19) No(6)
• 6C2=15 .
  
• 10 years agoHelpfull: Yes(16) No(12)
• i think ans is 12 ways
  it is given that atleast one of them should get exatly one object
  so take 3 places _ _ _
  fix 1 at 1st place 1 0 6-------3ways
  1 1 5-------3ways
  1 2 4-------3ways
  1 3 3-------3ways
  so totaly 3+3+3+3=12 ways
• 10 years agoHelpfull: Yes(9) No(18)
• 6C2=15 is correct one
• 10 years agoHelpfull: Yes(7) No(1)
• answer is :18 ways
  first possible way is 6,0,1
  this can be arranged in 3! ways so it is 6;
  second possible ways is 1,1,5 so it can be arranged in 3!/2! ways so it is 3;
  third way to arrang this 1,2,4 so it can be arranged in 3! ways;
  forth way to arrange this 1,3,3 so it is 3!/2! =3;
  so 3+6+6+3=18;
• 10 years agoHelpfull: Yes(5) No(6)
• Division of m+n+p objects into three groups is given by $displaystylefrac{{(m + n + p)!}}{{m! times n! times p!}}$ But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5 So The number of ways are $displaystyledisplaystylefrac{{(7)!}}{{1! times 3! times 3!}} times displaystylefrac{1}{{2!}} + displaystylefrac{{(7)!}}{{1! times 2! times 4!}} + displaystylefrac{{(7)!}} {{1! times 1! times 5!}} times displaystylefrac{1}{{2!}}$ = 70 + 105 + 21 = 196
• 10 years agoHelpfull: Yes(4) No(2)
• Possible Cases are:
  1 1 5 = 7 x 6
  1 2 4 = 7 x 6C2
  1 3 3 = 7 x 6C3
  1 0 6 = 7 x 6C0
  and their own arrangements are 3! i.e, 1st person get 1 object, 2nd get 0 object and 3rd get 6 object or 1st person get 0 object, 2nd get 1 object and 3rd get 6 object and so on..
  So total arrangement = ( 7 x 6 + 7 x 6C2 + 7 x 6C3 + 7 x 6C0 ) x 3!
  = 1764
• 10 years agoHelpfull: Yes(3) No(1)
• Atleast one person should have exactly one object.
  so 1 0 6= 3P3=6ways
  similarly
  1 1 5=3P3/2!=3ways
  1 2 4=3P3=6ways
  1 3 3=3P3/2!=3ways
  
  6+3+6+3=18
  
• 10 years agoHelpfull: Yes(2) No(5)
• i guess ans is 36
  =(7+3-1)C(3-1)
  =9C2
  =36
• 10 years agoHelpfull: Yes(2) No(3)
• each get atleast one object
  -> give one object to each of them
  ->can be done in 3 ways
  ->now divide remaining 4 object ->4*3*2 ways
  ->total 3*4*3*2 ways
• 10 years agoHelpfull: Yes(0) No(9)
• Sorry for last answer it was wrong:
  correct ans is 21*36;
  there is seven different things so ways of distributing of this things for only one thing 7*3=21
  now remaining is 2 person for which 6*6 ways =36;
  so total no of ways is 36*21;
• 10 years agoHelpfull: Yes(0) No(9)
• as we can select one from 3 in 3C1 ways and can assign him in 7C1 ways. rest each object can be given in two ways. So answer should be
  3C1*7C1*2*2*2*2*2*2
• 10 years agoHelpfull: Yes(0) No(4)
• 6C2*2!+5C1=35
• 10 years agoHelpfull: Yes(0) No(4)
• ans = 4C3 * 3! = 24 ??????
• 10 years agoHelpfull: Yes(0) No(2)
• One person must get exctly one..thrfore it can be given in 3 ways(to any one of d 3 person).. remaining can be given in 6p2 ways...(not 6c2). so ans wil b 3*6p2=90
• 10 years agoHelpfull: Yes(0) No(2)
• Suppose everyone gets at least one object, hence (7-3)=4 objects remain.
  The rest of the objects can be distributed in (4+3-1)C(3-1)=6C2=15 number of ways.
• 10 years agoHelpfull: Yes(0) No(2)
• seven objects must be divided among three persons
  so, we consider 7 objects (1,2,3,4,5,6,7)
  consider 3persons m,n,p
  m+n+p/m!n!p!
  possible ways for (1,1,5),(1,2,4),(1,3,3)
  (m+n+p)!=7!
  (7!/1!1!5!)1/2+(7!/1!2!4!)*1/2+7!/1!3!3!(1/2 means 1!1! combination is 1 so divide that value)
  ans=196
  
• 8 years agoHelpfull: Yes(0) No(0)
• first 3 objs are distributed among 3 persons remaining 4 objects can be distributed among 3 in 4c3*3! ways
  so 24 ways
• 8 years agoHelpfull: Yes(0) No(0)