Example Card: Permutations vs. Combinations

When choosing a group, does the order of selection change the outcome? Let's explore why sometimes it doesn't. This example highlights the key idea of how combinations relate to permutations.

Example Problem: A club has 5 members. How many distinct committees of 3 members can be formed?

1. First, let's find the number of ways to arrange 3 members out of 5. This is a permutation. $$ 5P 3 = \frac{5!}{(5 3)!} = \frac{5!}{2!} = 60 $$ There are 60 ordered arrangements. 2. However, for a committee, the order doesn't matter. A committee with members {A, B, C} is the same regardless of who was picked first. 3. For any group of 3 members, there are $3!$ ways to arrange them ($3 \cdot 2 \cdot 1 = 6$). These 6 arrangements all represent the same single committee. 4. To find the number of unique committees (combinations), we must divide the total permutations by the number of arrangements for each group. $$ \text{Combinations} = \frac{\text{Permutations}}{\text{Ways to order 3 members}} = \frac{ 5P 3}{3!} $$ 5. Calculating this gives us the answer: $$ \frac{60}{6} = 10 $$ There are 10 possible committees.