1. The problem is to find the number of ways to choose 10 items from a set of $n$ items, which is a combination problem. 2. The formula for combinations is given by: $$ C(n, k) = \frac{n!}{k!(n-k)!} $$ where $n$ is the total number of items, $k$ is the number of items to choose, and $!$ denotes factorial. 3. Important rules: - Order does not matter in combinations. - Factorials grow very fast, so simplify before calculating. 4. For example, if $n=15$ and $k=10$, then: $$ C(15, 10) = \frac{15!}{10!5!} $$ 5. Simplify the factorial expression by canceling common terms: $$ \frac{15 \times 14 \times 13 \times 12 \times 11 \times \cancel{10!}}{\cancel{10!} \times 5 \times 4 \times 3 \times 2 \times 1} $$ 6. Calculate numerator and denominator: $$ \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} $$ 7. Compute step-by-step: - Numerator: $15 \times 14 = 210$, $210 \times 13 = 2730$, $2730 \times 12 = 32760$, $32760 \times 11 = 360360$ - Denominator: $5 \times 4 = 20$, $20 \times 3 = 60$, $60 \times 2 = 120$, $120 \times 1 = 120$ 8. Final calculation: $$ \frac{360360}{120} = 3003 $$ 9. Therefore, the number of ways to choose 10 items from 15 is $3003$. 10. You can apply this method to any $n$ choose 10 problem by substituting the value of $n$.