Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, where order does not matter, is given by ${}^nC_r = \frac{n!}{r!(n-r)!}$.
• Division into Distinct Groups: The number of ways to divide $n$ distinct objects into $k$ distinct groups of sizes $n_1, n_2, \dots, n_k$ (where $n_1 + n_2 + \dots + n_k = n$) is $\frac{n!}{n_1! n_2! \dots n_k!}$.

Step-by-Step Solution

Step 1: Understanding the Problem

• Why: We need to partition the set $S = \{1, 2, 3, \dots, 12\}$ into three distinct sets A, B, and C, each containing 4 elements. The order in which the elements are chosen for each set matters because the sets are labelled A, B, and C.
• Math: $n = 12$, $k = 3$, $n_1 = n_2 = n_3 = 4$.

Step 2: Forming Set A

• Why: We start by selecting 4 elements out of 12 to form set A. Since the order of selection within set A doesn't matter, we use combinations.
• Math: The number of ways to choose 4 elements for set A from 12 is ${}^{12}C_4$. ${}^{12}C_4 = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!}$

Step 3: Forming Set B

• Why: After selecting elements for set A, we are left with 8 elements. We now select 4 elements from these remaining 8 to form set B.
• Math: The number of ways to choose 4 elements for set B from the remaining 8 is ${}^{8}C_4$. ${}^{8}C_4 = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!}$

Step 4: Forming Set C

• Why: After selecting elements for sets A and B, we are left with 4 elements. These remaining 4 elements must form set C.
• Math: The number of ways to choose 4 elements for set C from the remaining 4 is ${}^{4}C_4$. ${}^{4}C_4 = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1$

Step 5: Calculating the Total Number of Partitions

• Why: Since we are forming the sets sequentially, we multiply the number of ways to form each set to get the total number of ways to partition S into A, B, and C.
• Math: Total number of ways = ${}^{12}C_4 \times {}^{8}C_4 \times {}^{4}C_4$ $\frac{12!}{4!8!} \times \frac{8!}{4!4!} \times \frac{4!}{4!0!} = \frac{12!}{4!8!} \times \frac{8!}{4!4!} \times 1 = \frac{12!}{4!4!4!} = \frac{12!}{(4!)^3}$

Common Mistakes & Tips

• Labelled vs. Unlabelled Groups: Remember to consider whether the groups are labelled or unlabelled. If the groups were unlabelled, you would need to divide by $3!$ because the order in which the groups are formed wouldn't matter. In this case, the groups A, B, and C are labelled, so we do not divide by $3!$.
• Applying the Formula Directly: Recognizing that this is a direct application of the formula for dividing distinct items into distinct groups can save time.

Summary

We are partitioning a set of 12 distinct elements into three labelled sets of 4 elements each. We calculated the number of ways to do this by sequentially selecting elements for each set and multiplying the number of possibilities for each step. This gives us a total of $\frac{12!}{(4!)^3}$ ways.

Final Answer

The final answer is $\boxed{\frac{12!}{(4!)^3}}$, which corresponds to option (A).