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

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 :-
$\frac{(2+3-1)!}{2!(3-1)!}=6\left(\therefore \frac{(n+r-1)!}{n!(r-1)!}\right)$
(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.