Question

5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. Number of ways in which the balls can be placed so that no box remains empty, if :

Column I		Column II
A	balls are identical but boxes are different	p	2
B	balls are different but boxes are identical	q	2 5
C	balls as well as boxes are identical	r	5 0
D	balls as well as boxes are identical but boxes are kept in a row	s	6

• A. $( A ) \rightarrow( s ),( B ) \rightarrow( q ),( C ) \rightarrow( p ),( D ) \rightarrow( s )$
• B. $( A ) \rightarrow( p ),( B ) \rightarrow( p),( C ) \rightarrow( r ),( D ) \rightarrow( s )$
• C. $( A ) \rightarrow( q ),( B ) \rightarrow( r ),( C ) \rightarrow( p ),( D ) \rightarrow( s )$
• D. $( A ) \rightarrow(r ),( B ) \rightarrow( p ),( C ) \rightarrow( s ),( D ) \rightarrow( q )$

Answer 1

Answer: (A)

(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 :-
$\frac{(2+3-1) !}{2 !(3-1) !}=6 \quad\left(\therefore \frac{(n+r-1) !}{n !(r-1) !}\right)$
(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 : $\frac{5 !}{(1 !)(2 !)^2 \times 2 !}=15$
Case 2 : Division can also be 3 in one box and 1 each in remaining 2 boxes $(3,1,1)$
ways: $\frac{5 !}{3 ! \times(1 !)^2 \times 2 !}=10$
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 cases as that of in part B with arrangements.