1. The problem is to find the number of combinations of 3 items chosen from a set of $n$ items. 2. The formula for combinations is given by: $$ C(n, r) = \frac{n!}{r!(n-r)!} $$ where $n$ is the total number of items, $r$ is the number of items to choose, and $!$ denotes factorial. 3. For this problem, $r=3$, so the formula becomes: $$ C(n, 3) = \frac{n!}{3!(n-3)!} $$ 4. This formula counts the number of ways to choose 3 items from $n$ without regard to order. 5. To simplify, note that: $$ n! = n \times (n-1) \times (n-2) \times (n-3)! $$ 6. Substitute into the formula: $$ C(n, 3) = \frac{n \times (n-1) \times (n-2) \times (n-3)!}{3! \times (n-3)!} $$ 7. Cancel the common factor $(n-3)!$: $$ C(n, 3) = \frac{n \times (n-1) \times (n-2) \times \cancel{(n-3)!}}{3! \times \cancel{(n-3)!}} $$ 8. Since $3! = 3 \times 2 \times 1 = 6$, the formula simplifies to: $$ C(n, 3) = \frac{n \times (n-1) \times (n-2)}{6} $$ 9. This is the final formula to calculate the number of combinations of 3 items from $n$. 10. To find the exact number, substitute the value of $n$ into the formula and compute. Example: If $n=5$, then $$ C(5, 3) = \frac{5 \times 4 \times 3}{6} = \frac{60}{6} = 10 $$ This means there are 10 ways to choose 3 items from 5.