IN how many ways we can distribute 10 identical looking pencils to 4 students so that each student get at least one pencil?
a)5040
b)210
c)84
d)none of these

• 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, c is correct answer
• 12 years agoHelpfull: Yes(78) No(12)
• we have formulas regarding such sort of distributions:
  x1+x2+x3......xn=r, then 1)the number of +ve integral solutions(excluding zero)is (n-1)c(r-1) and 2)the number of -ve integral solutions (including zero) is (n+r-1)c(r-1)
• 12 years agoHelpfull: Yes(24) No(4)
• a1+a2+a3+....+ar=n
  1) No. of non negative integral solutions = n+r-1 C r-1
  2) No. of positive integral solutions : n-1 C r-1
  so in given question
  we have r=4(students having pencils)
  sum of all the pencils of students is 10
  so
  a1+a2+a3+a4=10
  No. of positive integral solutions : n-1 C r-1
  (as each student have atleast 1 pencil)
  10-1 C 4-1= 36
  
  
• 12 years agoHelpfull: Yes(10) No(26)
• firstly give 1 pencil each to the 4, now we can distribute the remaining 6 pencils any way we like.
  using the "stars and bars" formula, (6+4-1)C(4-1)
  = 9C3 = 84 [ c ]
• 11 years agoHelpfull: Yes(9) No(0)
• (n+r-1)c(r-1)
• 12 years agoHelpfull: Yes(7) No(4)
• @Ragini: Plzz explain how this come:
  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
  How u GOT (6+4-1)C(4-1) ???
  Thx in Advance !!!! :) :)
• 12 years agoHelpfull: Yes(6) No(4)
• ragini ji,plz elaborate it....
• 12 years agoHelpfull: Yes(5) No(3)
• 10C4=210
  Out of 10 pencils 4 are given to student each
• 12 years agoHelpfull: Yes(4) No(67)
• answer=84
  Number of ways in which n identical things can be distributed among r persons, each one of whom can receive 1,2 or more items (n-1)C(r-1).
  so answer =9c3=84,c is correct
• 10 years agoHelpfull: Yes(4) No(2)
• ans 84
  n-1 c r-1 as at least one so non of them get zero
• 9 years agoHelpfull: Yes(3) No(0)
• The no. of ways in which 'n' identical things can be divided in 'r' groups when each person get any no. of thing including zero= (n+r-1)C(r-1)
  And
  The no. of ways in which 'n' identical things can be divided in 'r' groups when each person get atleast one thing = (n-1)C(r-1)
  
  
  
  (10-1) C (4-1)= 9! / ( 6! * 3! ) = 84. c is the correct answer.
• 9 years agoHelpfull: Yes(3) No(0)
• Ans:b)210
  Solution:
  =(10*9*8*7)/(4*3*2*1)
  =210
• 12 years agoHelpfull: Yes(2) No(49)
• qdssdfdsfadsf
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• 11 years agoHelpfull: Yes(2) No(9)
• the formula is (n-1)C(r-1)= 9C3= 84
• 9 years agoHelpfull: Yes(2) No(0)