Question

Number of ways in which four distinct balls can be put into two identical boxes so that no box remains empty, is

• A. $5$
• B. $1$
• C. $7$
• D. $3$

Answer 1

The correct option is C $7$

We can divide the balls into two groups : $(1,3)$ and $(2,2)$
In $(2,2),$ both groups have equal number of balls.
Now, number of ways
$=\frac{4!}{1!3!}+\frac{4!}{2!2!}\times \frac{1}{2!}$
$=4+3=7$