A.
Each box (say B
1
,B
2
,B
3
) will have at least one ball.
Now the ways for placing other 2 identical balls in 3 different boxes are
2!(3−1)!
(2+3−1)!
=6(∴
n!(r−1)!
(n+r−1)!
)
B.
Case 1 : 5 balls can be divided in 3 groups having 2 balls each in 2 boxes and 1 ball for in third box (2,2,1)
ways :
(1!)(2!)
2
×2!
5!
=15
Case 2 : Division can also be 3 in one box and 1 each in remaining 2 boxes (3,1,1)
ways :
(3!)(1!)
2
×2!
5!
=10
Hence total ways =10+15=25
C.
Only 2 arrangements are possible.
1. 2 balls each in 2 boxes & remaining ball in other box (2,2,1)
2. 3 balls in 1 box and 1 ball each in other boxes (3,1,1)
D.
Same as part A
