Five balls are placed in three boxes. Each box can hold all the five balls. In how many ways can we place the balls in the boxes so that no box remains empty if
List - 1 List-II
i) balls are identical and the a) 25
boxes are different
ii) balls are different but b) 6
boxes are identical
iii) the balls as well as
boxes are identical c) 150
d) 2
The correct Match is

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b, c, d

c, b, d

d, a, b

b, a, d

answer is B.

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Detailed Solution

i) 5 identical balls to be distributed into 3 different boxes so that no box is empty.

(3,1,1),(1,1,3),(1,3,1)(1,2,2),(2,1,2),(2,2,1)

number of ways=6

(ii)5 different balls to be distributed to 3 identical boxes

number of ways of dividing 5 balls into (3,1,1) and (2,2,1)= $\frac{5!}{3! {(1!)}^{2} 2!} + \frac{5!}{(2! 2! 1!) 2!} = 10 + 15 = 25$

(iii)5 identical balls to be distributed into 3 identical boxes

(3,1,1),(2,2,1)

Number of ways=2

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