In how many ways 10 identical pencils can be divided among 4 so that each gets atleast 1 pencil.

• div of n identical element into r groups such that each group contain atleast 1 element= n-1Cr-1
  10-1C4-1= 9C3=84
• 11 years agoHelpfull: Yes(33) No(1)
• 9c3=84 ans
• 11 years agoHelpfull: Yes(6) No(1)
• n-1Cr-1=10-1C4-1=9C3=84
  in case:at least one we have to deduct 0 case...=1
  ans=84-1=83
• 11 years agoHelpfull: Yes(5) No(0)
• consider 4 students as A,B,C,D THEN A+B+C+D=10, but each one should get atleast 1 pencil so giving 1 pencil to each student, the remaining pencils are 10-4=6
  now A'+B'+C'+D'=6, number of non negative integers of A',B',C',D' ARE (6+4-1)C(4-1) I.E COMBINATIONS, 9c3=84
• 11 years agoHelpfull: Yes(4) No(0)
• m+n=8
  m-n=6
  solving simultn.v get m=7,n=1
  mn=7*1=7
  mn/6
  7/6=1 remainder
• 11 years agoHelpfull: Yes(3) No(0)
• first distribute each pencil into 4 then 6 left.
  after that its like arranging 6 identical pencils to 4
  6+4-1C4-1
• 11 years agoHelpfull: Yes(1) No(1)
• 9c3
  because the formula for n identical items dividing among r people such that at least each can get one item is n-1Cr-1
  
  
• 11 years agoHelpfull: Yes(1) No(0)
• here it is given atleast 1, can we apply (n-1)c(r-1)????
• 11 years agoHelpfull: Yes(0) No(0)
• 10+4-1c4-1= 286 including 0
  but excluding 0 it is 285
  am i wright or wrong plz tell me......
• 11 years agoHelpfull: Yes(0) No(4)