Answer the questions below. \( \begin{array}{l}\text { (a) A pizza parlor has a choice of } 11 \text { toppings for its pizzas. From these } 11 \text { toppings, how many } \\ \text { different 7-topping pizzas are possible? } \\ \square \\ \text { (b) A company that makes crayons is trying to decide which } 3 \text { colors to include in a promotional } \\ \text { mini-box of } 3 \text { crayons. The company can choose the } 3 \text { mini-box colors from its collection of } 70 \\ \text { colors. How many mini-boxes are possible? } \\ \square\end{array} \)

Certainly! Let's dive into some fun ways to solve these combination problems! First, when it comes to creating a 7-topping pizza from 11 different toppings, we're looking to choose 7 out of those 11 without caring about the order. This is a classic combination problem! The formula to calculate combinations is given by: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] So for the pizza, we calculate: \[ \binom{11}{7} = \frac{11!}{7! \cdot 4!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \] Thus, the number of different 7-topping pizzas possible is **330**. Now for the crayon box problem, we're also looking at combinations since the order in which we choose the colors doesn’t matter. Using the same combinations formula, we compute: \[ \binom{70}{3} = \frac{70!}{3! \cdot 67!} = \frac{70 \times 69 \times 68}{3 \times 2 \times 1} = 52360 \] Therefore, the number of different mini-boxes of crayons possible is **52,360**. And there you have it—two tasty equations solved! 🍕🖍️