10 nos of different gifts have to distributed among 4 people so that every one get at-least 1 gifts..

• (10-1)C(4-1)= 84
• 10 years agoHelpfull: Yes(36) No(4)
• as the total is 10 no. of gifts different possibilities are
  (7,1,1,1) = 4!/3! = 4
  (6,2,1,1) = 4!/2! =12
  (5,3,1,1) = 4!/2! =12
  (5,2,2,1) = 4!/2! =12
  (4,4,1,1) = 4!/(2!*2!) = 6
  (4,3,2,1) = 4! =24
  (4,2,2,2) = 4!/3! = 4
  (3,3,3,1) = 4!/3! = 4
  (3,3,2,2) = 4!/(2!*2!) = 6
  
  total possibilities is the addition of the above = 84
• 10 years agoHelpfull: Yes(23) No(5)
• as gifts r distinct so each price can b
  given in 4 ways so ans will b 4^10 and excluding d condition of getting 0 gifts. ans is 4^10-1
• 10 years agoHelpfull: Yes(15) No(1)
• (10+4-1)C(4-1)=13C3
• 10 years agoHelpfull: Yes(12) No(4)
• for distributing n identical thing among r people,so that each get atleast 1 is
  n-1Cr-1
  9C3=42
• 10 years agoHelpfull: Yes(6) No(17)
• in such hind of question use the formula n-1cr-1
• 10 years agoHelpfull: Yes(5) No(0)
• the gifts are different so we can arrange in 4^10 ways ,
  but due to the restriction that every one get at-least one.
  select 4 from 10 diff. gifts in 10C4 way and we can arrange this in 4! way.
  The remaining gifts are distributed in 4^6 ways without restriction.
  hence the naser will be : 10C4*4!*4^6.
• 10 years agoHelpfull: Yes(4) No(1)
• 10C4 is the correct answer.9C3 will be correct if and only if it was identical gifts.
• 10 years agoHelpfull: Yes(4) No(1)
• using partition rule,
  when each people has to have atleast one thing.
  (n-1)C(r-1)
  therefore,
  9C3=84
• 10 years agoHelpfull: Yes(4) No(1)
• (n-1)c(r-1) 9c3 =84ans
• 10 years agoHelpfull: Yes(2) No(1)
• (n-1)c(r-1) 9c3 =84
• 10 years agoHelpfull: Yes(2) No(2)
• since everyone sud get atleast 1 gift
  so
  s+x+y+z=10-(4)
  s+x+y+z=6
  (6+4-1)c(4-1)=84
• 10 years agoHelpfull: Yes(1) No(0)
• formula is (n-k)c(r-k)...n->total object
  r->here stand for no of people
  k-> k no of object always get choosen
  
  so (10-1)c(4-1)=9C3=84
• 10 years agoHelpfull: Yes(1) No(0)
• 13c4=13!/9!4!
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• Since the question is to distribute different gifts so 4 gifts can be selected from 10 gifts in 10C4 ways such that each has atleast 1 gift and this can be arranged in 4 ways Now 6 gifts are left to be distributed and this can be distributed inwhen only 1 gets all the 6 gifts- 4 wayswhen two gets- 6c51c16c42c26c33c362when three of them get- 6c42c11c16c33c21c16c24c22c264when all 4 of them get-6c33c12c11c16c24c22c11c124now summing up all the types of distribution 10c4446c51c16c42c26c33c3626c42c11c16c33c21c16c24c22c2646c33c12c11c16c24c22c11c124
• 10 years agoHelpfull: Yes(0) No(1)
• ANS IS 1
  55/3=1
  55/3^2=1
  55/3^3=1
• 10 years agoHelpfull: Yes(0) No(1)
• 4*(6+5+4+3+2+1)=84
  
• 9 years agoHelpfull: Yes(0) No(0)