1. The problem is to find the number of ways to choose 3 items from a set of $n$ items using combinations. 2. The formula for combinations (choosing $k$ items from $n$ without regard to order) is: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n!$ denotes the factorial of $n$. 3. For $k=3$, the formula becomes: $$\binom{n}{3} = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{3 \times 2 \times 1} = \frac{n(n-1)(n-2)}{6}$$ 4. This formula counts the number of unique groups of 3 items that can be selected from $n$ items. 5. To use this formula, substitute the value of $n$ (the total number of items) into the expression and simplify. 6. For example, if $n=5$, then: $$\binom{5}{3} = \frac{5 \times 4 \times 3}{6} = \frac{60}{6} = 10$$ 7. So, there are 10 ways to choose 3 items from 5. This method is called the combination method or "choose" method because it counts the number of ways to choose items without regard to order.