Question

The number of ways in which we can put $n$ distinct things in two identical boxes so that no box is empty, is

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

Answer 1

Answer: (C)

$2^{n-1}-1$

Here,the object are different number of objects $=n$
The objects are distributed into two $(r=2)$ boxes.
Since boxes are identical, so order is not to be considered.
No box remain empty (no blank lots)
Using the formula, the number of ways
$=\frac{1}{r!}[r^n{-}^r{C}_1{(r-1)}^n+...{(-1)}^{r-1}{.}^r{C}_{r-1}]$
where $r=2,n=2$
Hence, the required number of ways are
$=\frac{2^n{-}^2{C}_1{(2-1)}^n{+}^2{c}_2{(2-2)}^n}{2!}$

$=\frac{2^n-2}{2}$

$=2^{n-1}-1$