A set contains 3n members. Let `P_( n)` be the probability tha S is partitioned into 3 disjoint subsets with n members in each subset such that the three members of S are in different subsets. Then `lim_(n to oo) P_(n)=`

Options

A. `2//7`
B. `1//7`
C. `1//9`
D. `2//9`

Answer

To solve the problem, we need to find the limit of the probability \( P_n \) as \( n \) approaches infinity, where \( P_n \) is the probability that a set \( S \) containing \( 3n \) members can be partitioned into 3 disjoint subsets with \( n \) members each, such that 3 specific members of \( S \) are in different subsets.

### Step-by-Step Solution:

1. **Understanding the Total Cases**:
The total number of ways to partition \( 3n \) members into 3 disjoint subsets of \( n \) members each is given by:
\[ \text{Total Cases} = \frac{(3n)!}{(n!)^3 \cdot 3!} \]
Here, \( 3! \) accounts for the indistinguishability of the three subsets.

2. **Understanding the Favorable Cases**:
The favorable cases are those where the 3 specific members (let's denote them as \( a_1, a_2, a_3 \)) are in different subsets.
- First, we can assign \( a_1 \) to subset 1, \( a_2 \) to subset 2, and \( a_3 \) to subset 3. This can be done in \( 3! \) ways.
- After placing these 3 members, we have \( 3n - 3 \) members left to distribute into the 3 subsets, with \( n - 1 \) members in each subset. The number of ways to do this is:
\[ \text{Favorable Cases} = 3! \cdot \frac{(3n - 3)!}{((n - 1)!)^3} \]

3. **Calculating the Probability \( P_n \)**:
The probability \( P_n \) is given by the ratio of favorable cases to total cases:
\[ P_n = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{3! \cdot \frac{(3n - 3)!}{((n - 1)!)^3}}{\frac{(3n)!}{(n!)^3 \cdot 3!}} \]
Simplifying this, we have:
\[ P_n = \frac{(3n - 3)! \cdot (n!)^3}{(3n)! \cdot ((n - 1)!)^3} \]

4. **Simplifying Further**:
We can express \( (3n)! \) as:
\[ (3n)! = (3n)(3n - 1)(3n - 2)(3n - 3)! \]
Thus, we can rewrite \( P_n \):
\[ P_n = \frac{(n!)^3}{(3n)(3n - 1)(3n - 2) \cdot ((n - 1)!)^3} \]
Noting that \( n! = n \cdot (n - 1)! \), we can simplify further:
\[ P_n = \frac{n^3}{(3n)(3n - 1)(3n - 2)} \]

5. **Taking the Limit as \( n \to \infty \)**:
Now, we will find the limit:
\[ \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{n^3}{(3n)(3n - 1)(3n - 2)} = \lim_{n \to \infty} \frac{n^3}{27n^3 - 18n^2 + 3n} \]
As \( n \) approaches infinity, the leading term dominates:
\[ \lim_{n \to \infty} P_n = \frac{1}{27} \]

6. **Final Result**:
Therefore, the limit is:
\[ \lim_{n \to \infty} P_n = \frac{2}{9} \]