Question

The number of ways in which 5 distinct balls can be partitioned into exactly 2 non-empty subsets is

• A. 7
• B. 8
• C. 15
• D. 16

Answer 1

Answer: (C)

15

Answer 2

This problem asks for the number of ways to partition 5 distinct balls into exactly 2 non-empty subsets. This is a classic problem solved using Stirling numbers of the second kind, denoted as $S(n, k)$, where $n$ is the number of distinct objects and $k$ is the number of non-empty subsets.

We need to calculate $S(5, 2)$. The formula for Stirling numbers of the second kind is: $S(n, k) = \frac{1}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^n$

For $S(5, 2)$: $S(5, 2) = \frac{1}{2!} \left( (-1)^{2-0} \binom{2}{0} 0^5 + (-1)^{2-1} \binom{2}{1} 1^5 + (-1)^{2-2} \binom{2}{2} 2^5 \right)$ $S(5, 2) = \frac{1}{2} \left( (1)(1)(0) + (-1)(2)(1) + (1)(1)(32) \right)$ $S(5, 2) = \frac{1}{2} (0 - 2 + 32)$ $S(5, 2) = \frac{30}{2} = 15$

Alternatively, we can consider the ways to form two pairs from 5 distinct balls. This implies one ball will be left alone. The number of ways to choose the first pair is $\binom{5}{2}$. The number of ways to choose the second pair from the remaining 3 balls is $\binom{3}{2}$. Since the order of the two pairs does not matter, we divide by $2!$. Number of ways = $\frac{\binom{5}{2} \binom{3}{2}}{2!} = \frac{10 \times 3}{2} = 15$.

Both interpretations lead to 15.