The set S={1,2,3,...,12} is to be partitioned into three sets, A, B, C of equal size. ThusA∪B∪C=S,A∩B=B∩C=A∩C=ϕ. The number of ways to partition S is

Options

A. 1
B. 2
C. 3
D. more than 3

Answer

To solve the problem of partitioning the set \( S = \{1, 2, 3, \ldots, 12\} \) into three subsets \( A, B, C \) of equal size, we will follow these steps:

### Step 1: Determine the size of each subset
Since the set \( S \) has 12 elements and needs to be divided into three equal subsets, each subset will contain:
\[ \text{Size of each subset} = \frac{12}{3} = 4 \]

**Hint:** Remember that when partitioning a set into equal parts, the size of each part is the total number of elements divided by the number of parts.

### Step 2: Calculate the number of ways to choose elements for the first subset
We can choose 4 elements from the 12 elements for subset \( A \). The number of ways to choose 4 elements from 12 is given by the combination formula \( \binom{n}{r} \):
\[ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4! \cdot 8!} \]

**Hint:** Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to find the number of ways to choose \( r \) elements from \( n \) elements.

### Step 3: Calculate the number of ways to choose elements for the second subset
After choosing 4 elements for subset \( A \), we have 8 elements left. We now choose 4 elements for subset \( B \) from these 8 elements:
\[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4! \cdot 4!} \]

**Hint:** After selecting elements for one subset, always remember to adjust the total number of elements available for the next selection.

### Step 4: Determine the third subset
The remaining 4 elements will automatically go to subset \( C \). There is only one way to assign these remaining elements to \( C \):
\[ \binom{4}{4} = 1 \]

**Hint:** When all remaining elements must go into one subset, there is only one way to do this.

### Step 5: Calculate the total number of ways to partition the set
The total number of ways to partition the set \( S \) into three subsets \( A, B, C \) is the product of the combinations calculated in the previous steps. However, since the order of subsets does not matter (i.e., \( A, B, C \) is the same as \( B, C, A \)), we must divide by the number of ways to arrange the three subsets, which is \( 3! \):
\[ \text{Total ways} = \frac{\binom{12}{4} \cdot \binom{8}{4} \cdot \binom{4}{4}}{3!} \]

### Step 6: Substitute the values and calculate
Now we substitute the values:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]
\[ \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]
\[ \binom{4}{4} = 1 \]
Thus, the total number of ways is:
\[ \text{Total ways} = \frac{495 \times 70 \times 1}{6} = \frac{34650}{6} = 5775 \]

### Final Answer
The number of ways to partition the set \( S \) into three subsets \( A, B, C \) of equal size is \( 5775 \).