Step 3: Count the number of distinct arrangements of the multiset {A, A, B, C}. The number of distinct permutations is:
["# Step 3: Counting Distinct Arrangements of the Multiset {A, A, B, C}", "When working with permutations of a multiset—a set that allows repeated elements—the total number of distinct arrangements (or permutations) must account for repeated items to avoid overcounting identical arrangements.", "In this example, we want to count how many distinct ways we can arrange the multiset {A, A, B, C}.", "---", "### What Is a Multiset?", "A multiset is a generalization of a set that allows multiple instances of elements. Here, the multiset consists of:\n- Two A’s (×2)\n- One B (×1)\n- One C (×1)", "---", "### Total Permutations of a Multiset", "For a multiset with total $ n $ elements and counts $ n_1, n_2, ..., n_k $ of each distinct element (such that $ n_1 + n_2 + ... + n_k = n $), the number of distinct permutations is given by:", "$$\n\frac{n!}{n_1! \ imes n_2! \ imes \cdots \ imes n_k!}\n$$", "Where:\n- $ n! $: factorial of total number of items\n- $ n_1!, n_2!, \ldots, n_k! $: factorials of counts of each repeated element", "---", "### Applying the Formula", "For the multiset {A, A, B, C}:\n- Total number of letters, $ n = 4 $\n- Two A’s → $ n_1 = 2 $\n- One B → $ n_2 = 1 $\n- One C → $ n_3 = 1 $", "So the number of distinct permutations is:", "$$\n\frac{4!}{2! \ imes 1! \ imes 1!} = \frac{24}{2 \ imes 1 \ imes 1} = \frac{24}{2} = 12\n$$", "---", "### Final Answer", "The number of distinct arrangements of the multiset {A, A, B, C} is:", "12", "---", "This method ensures you count only unique sequences, eliminating duplicates caused by identical letters. Understanding how to handle multisets is essential for solving combinatorics problems efficiently."]
