THE ARE 20 STUDENTS AND THEY NEED TO BE ALLOCATED INTO 3 CLASSROOMS.IN HOW MANY WAYS CAN THIS BE DONE
1.ASSUMING THAT THE STUDENTS ARE IDENTICAL
2.ASSUMING THAT THE STUDENTS ARE DISTINGUISHABLE

• 1. 20 students have to be allocate in 3 identical classrooms
  n=20,r=3
  no. of ways=(n+r-1)C(r-1)= (20+3-1)C(3-1)= 22C2 = 231
  
  2. 20 students have to be allocate in 3 DISTINGUISHABLE classroms
  each student can be allocated in 3 ways so total
  no. of ways = 3^20
• 10 years agoHelpfull: Yes(75) No(9)
• 1. For any identical related question, the ans. is 1(one)
  
  2. 3^20
• 10 years agoHelpfull: Yes(4) No(9)
• 1. Like the possible solutions of the integral equations
  x + y + z = 20 (z,y & z as three classrooms)
  Arrange r = 20 possble combinations among n=3 elements.
  There are H(n, r) = C(n+r-1, r) = C(22,20) = C(22,2)
  =22*21/2 = 21*11 = 231
  2. Each student has 3 choices, totally, there 3^20 ways.
• 10 years agoHelpfull: Yes(2) No(2)
• @NAOREM, with exclusion (no class can contain more than one student) then For any identical related question, the ans will be one.
• 10 years agoHelpfull: Yes(2) No(0)
• 1.given that students are identical so....
  
  it is 2^3-1=7
  
  2.given that students are diffrent then
  it can be done in n+r-1C3 ways 1.e 20+3-1 C 3-1 = 22C2 ways
• 10 years agoHelpfull: Yes(1) No(13)
• @Naorem is correct
• 10 years agoHelpfull: Yes(1) No(4)
• formula for the solutions:
  for identical case : (n-r-1)C(r-1)
  for distinguisale case: (n-r-1)P(r-1)
• 10 years agoHelpfull: Yes(0) No(9)