The set `S""=""{1,""2,""3,""........ ,""12)` is to be partitioned into three sets A, B, C of equal size. Thus, `AuuBuuC""=""S ,""AnnB""=""BnnC""=""AnnC""=varphi` . The number of ways to partition S is (1) `(12 !)/(3!(4!)^3)` (2) `(12 !)/(3!(3!)^4)` (3) `(12 !)/((4!)^3)` (4) `(12 !)/((4!)^4)`

Options

A. `(12!)/(3!(4!)^(3))`
B. `(12!)/(3!(3!)^(4))`
C. `(12!)/((4!)^(3))`
D. `(12!)/((4!)^(4))`

Explanation

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

### Step 1: Determine the size of each set
Since the total number of elements in set \( S \) is 12 and we want to partition it into three equal sets, each set will contain:
\[ \text{Size of each set} = \frac{12}{3} = 4 \]

### Step 2: Calculate the number of ways to select elements for set \( A \)
To select 4 elements for set \( A \) from the 12 elements, we use the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus, the number of ways to choose 4 elements for set \( A \) is:
\[ \text{Ways for } A = \binom{12}{4} = \frac{12!}{4! \cdot (12-4)!} = \frac{12!}{4! \cdot 8!} \]

### Step 3: Calculate the number of ways to select elements for set \( B \)
After selecting 4 elements for set \( A \), there are 8 elements left. We now select 4 elements for set \( B \):
\[ \text{Ways for } B = \binom{8}{4} = \frac{8!}{4! \cdot (8-4)!} = \frac{8!}{4! \cdot 4!} \]

### Step 4: Calculate the number of ways to select elements for set \( C \)
After selecting elements for sets \( A \) and \( B \), there are 4 elements remaining, which will all go to set \( C \):
\[ \text{Ways for } C = \binom{4}{4} = 1 \]

### Step 5: Combine the selections
The total number of ways to partition the set \( S \) into sets \( A, B, C \) is the product of the number of ways to select each set:
\[ \text{Total ways} = \text{Ways for } A \times \text{Ways for } B \times \text{Ways for } C \]
Substituting the values we calculated:
\[ \text{Total ways} = \binom{12}{4} \times \binom{8}{4} \times 1 = \frac{12!}{4! \cdot 8!} \times \frac{8!}{4! \cdot 4!} \times 1 \]

### Step 6: Simplify the expression
Notice that the \( 8! \) cancels out:
\[ \text{Total ways} = \frac{12!}{4! \cdot 4! \cdot 4!} \]

### Step 7: Account for the indistinguishability of sets
Since sets \( A, B, \) and \( C \) are indistinguishable, we must divide by the number of ways to arrange these three sets, which is \( 3! \):
\[ \text{Final count} = \frac{12!}{4! \cdot 4! \cdot 4! \cdot 3!} \]

### Final Answer
Thus, the number of ways to partition the set \( S \) is:
\[ \frac{12!}{3! \cdot (4!)^3} \]

### Conclusion
The correct option is (2) \( \frac{12!}{3! \cdot (4!)^3} \).

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