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case (3):Identical objectIdentical groupsAny way (No Restriction)Each group must containatleast one object (⩾1)

To solve these type of problems knowledge of generating functions will be required.

P(n, r) = P(n-r, r) + P(n-1, r-1)Recursive formoulaSpecial cases: -P(0, 0) = 1P(n, n) = 1P(n, 1) = 1P(n, 2) = if n is even = if n is oddP(n, n-1) = 1P(n, r) = 0 if r>n
case (3):Identical objectIdentical groupsAny way (No Restriction)Each group must containatleast one object (⩾1)(n)(r)P(n, r)

(Q). In how many ways 5 identical balls can be distributed to 10 identical bags if:
(i) There is no restriction
(ii) Each bag must contain atleast 1 ball
(iii) Each bag must contain at most 1 ball
(iv) Each bag must contain exactly 1 ball
Solution:-

Let first do it manually:-

5IdenticalIdentical1234567BBBBBBBBBB5433221111122111111100000000000000000000000000000000000000000000000000

Let’s take 3rd case:-
3 2 0 0 0 0 0 0 0 0
Here it does not matter that these 3 & 2 balls packet are given to which bag because bags are identical i.e. indistinguishable.
Also since balls are also identical so there is always 1 way to select any these 3 & 2 balls out of 5 balls.
So there is only 1 way to distribute this division.

Similar situation applies for all other cases.
Hence total number of ways to distribute = Answer

(ii) 0 (see diagram) Answer

(iii) 1 (7ᵗʰ case) Answer

(iv) 0 (see diagram) Answer

n = 5
r = 10

(i)= P(5, 1) + P(5, 2) + P(5, 3) + P(5, 4) + P(5, 5) + P(5, 6) + P(5, 7) + P(5, 8) + P(5, 9) + P(5, 10) 0 0 0 0 0 Since P(n, r) = 0 if r>n= P(5, 1) + P(5, 2) + P(5, 3) + P(5, 4) + P(5, 5) = 1 + + + 1 12now P(5, 3) = P(2, 3) + P(4, 2)0 = 2= 0 + 2 = 2Hence value of equation (i) will be = 1 + 2 + 2 + 1 + 1 = 7 Answer (ii) P(5, 10) = 0 Answerequation (i)

⦿ Now Let us take another example.
(Q). In how many ways 12 identical chocolates can be distributed to 5 identical bags:-
(i) Any way
(ii) Each bag contains atleast 1 chocolate
(iii) Each bag contains at most 1 chocolate
(iv) Each bag contains exactly 1 chocolate
Solution:-

 

(i) = P(12, 1) + P(12, 2) + P(12, 3) + P(12, 4) + P(12, 5) 1 6equation (i)⦿ now P(12, 3) = P(9, 3) + P(11, 2)= = 5= P(9, 3) + 5= P(6, 3) + P(8, 2) + 5 4= P(6, 3) + 9= P(3, 3) + P(5, 2) + 9= 1 + 2 + 9= 12⦿ now P(12, 4) = P(8, 4) + P(11, 3) = P(4, 4) + P(7, 3) + P(8, 3) + P(10, 2) = 1 + P(7, 3) + P(8, 3) + 5 = 6 + P(4, 3) + P(6, 2) + P(5, 3) + P(7, 2)1 3 3= 6 + 1 + 3 + P(5, 3) + 3= 13 + P(5, 3)= 13 + P(2, 3) + P(4, 2) = 13 + 0 + 2= 15⦿ now P(12, 5) = P(7, 5) + P(11, 4) = P(2, 5) + P(6, 4) + P(7, 4) + P(10, 3)0= P(2, 4) + P(5, 3) + P(3, 4) + P(6, 3) + P(7, 3) + P(9, 2)= 0 + P(5, 3) + 0 + P(6, 3) + P(7, 3) + 4= 4 + P(5, 3) + P(6, 3) + P(7, 3)= 4 + P(2, 3) + P(4, 2) + P(3, 3) + P(5, 2) + P(4, 3) + P(6, 2)= 4 + 0 + 2 + 1 + 2 + 1 + 3= 13 Hence value of equation (i) will be = 1 + 6 + 12 + 15 + 13 = 47 Answer

(ii) P(12, 5) = 13 Answer

(iii) 0 Answer

(iv) 0  Answer