1. The problem is to understand and calculate combinations, often denoted as $nCr$, which represent the number of ways to choose $r$ elements from a set of $n$ elements without regard to order. 2. The formula for combinations is: $$nCr = \frac{n!}{r!(n-r)!}$$ where $n!$ (n factorial) is the product of all positive integers up to $n$. 3. Important rules: - $n$ and $r$ must be non-negative integers. - $r$ cannot be greater than $n$. - Factorials grow very fast, so simplify before calculating. 4. Example: Calculate $5C2$. $$5C2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2! \times 3!} = \frac{5 \times 4}{2 \times 1} = 10$$ 5. Explanation: We cancel $3!$ in numerator and denominator, then multiply the remaining numbers in numerator and denominator and divide. 6. This means there are 10 ways to choose 2 elements from 5. Final answer: $nCr = \frac{n!}{r!(n-r)!}$ and example $5C2 = 10$.