First slide

Combinations

Question

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

Moderate

A

$2^{n} - 2$
B

$2^{n} - 1$
C

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

$n^{2} - 2$

Solution

Let us first label the boxes $B_{1}$ and $B_{2}$ Now, each
object can be put either in $B_{1}$ or in $B_{2}$ . So, there are two ways to
deal with each of then objects. Consequently, $n$ objects can be
dealt with $2^{n}$ ways. Out of these are $2^{n}$ ways. there are two
ways (i) when all objects are put in box $B_{1}$ (ii) when all objects
are put in box $B_{2}$ . Thus, there $2^{n} - 2$ ways in which neither box is empty. If we now remove the labels from the boxes so that they become identical, this number must be divided by 2 to get the required number of ways.
$∴$ Required number of ways $= \frac{1}{2} (2^{n} - 2) = 2^{n - 1} - 1$ .

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