in how man ways can 10 identical presents be distributed among 6 children so that each gets at least one present.
a)15c5
b)16c6
c)9c5
d)6 ^ 10

• No of ways of dividing n identical things among r different group(for non empty set i.e, each one get at least one) is given by (n-1)C(r-1)
  so on applying this to given situation we get. 9c5
• 9 years agoHelpfull: Yes(25) No(2)
• ans is 9C5 bcz in question they have mentioned that each child get atleast one present so, first distribute 6 presents to 6 childrens remaining is 4 presents(n=4) now apply formula (n+r-1)C(r-1) (4+6-1)C(4-1) =9C5
  
  (n+r-1)C(r-1) is used when n identical things are distributed to r people where each can get either 0 or more presents
• 9 years agoHelpfull: Yes(5) No(0)
• totel=10
  remaing r=10-6=4
  n+r-1Cr+1= 6+4-1C4+1
  9C5
• 9 years agoHelpfull: Yes(2) No(2)
• since presents are identical
  can be distributed in 6^10 ways
• 9 years agoHelpfull: Yes(2) No(5)
• sorry but answer should be like dividing n identical things among r different group=(n+r-1)C(r-1)
  which will come out as 15C5
  so (a) correct choice
  
• 9 years agoHelpfull: Yes(1) No(3)
• here n=10 & r= 6
  we want to distribute identical presents among 6 children each gets at least one present n-1cr-1.
  9c5 is answer
• 9 years agoHelpfull: Yes(1) No(0)
• ans- 9c5
  concept is find the no. of positive integeral solutions of equation- x1 + x2 + x3 + x4 + x5 + x6 = 10
  is-9c5 [ (10-1) c (6-1) ]
  
• 9 years agoHelpfull: Yes(0) No(0)
• consider it like this....
  x+x+x+x+x+x = 10
  ie each of them should receive atleat one and could receive more than one...
  (n-1)C(r-1)
• 9 years agoHelpfull: Yes(0) No(0)
• it is a situation of distributon of n identical objetc among r groups such that every one must get atleast one object=>
  "n-1Cr-1"
  => (10-1) C(6-1)
  => 9C5
• 8 years agoHelpfull: Yes(0) No(0)
• 9c5 answer
• 5 years agoHelpfull: Yes(0) No(0)