Question

The number of ways in which n distinct objects can be put into two identical boxes so that no box remains empty is

• A. $2^n-2$
• B. $2^n-1$
• C. $2^{n-1}-1$
• D. $n^2-2$

Answer 1

Answer: (C)

$2^{n-1}-1$

Since there are $2$ boxes, $n$ objects can be placed in $2^n$ ays.
Now both the boxes can be arranged in $2!$ ....(identical boxes)
Hence, we get
$\frac{2^n}{2!}$
$=2^{n-1}$
However this also includes the case where all objects goes into one box..
Hence, the required permutation is
$=2^{n-1}-1$