91Ó°ÊÓ

We are given: - 3 red marbles (R) - 2 green marbles (G) - 1 lavender marble (L) - 2 yellow marbles (Y) - 2 orange marbles (O) And we want to find the number of sets of five marbles that include at least two red marbles.

We need to find the number of ways to choose five marbles from the bag, with at least two red ones. There are three possible scenarios: 1. Choose exactly 2 red marbles and 3 marbles from the other colors. 2. Choose exactly 3 red marbles and 2 marbles from the other colors. 3. Choose all 5 red marbles.

Now we will count the number of combinations for each scenario using the combination formula: \[C(n, k) = \frac{n!}{k!(n-k)!}\] 1. Choosing exactly 2 red marbles and 3 marbles from the other colors: - Choose 2 red marbles from 3: \(C(3, 2)\) - Choose 3 marbles from the remaining 7 of other colors: \(C(7, 3)\) - Total combinations for this scenario: \(C(3, 2) \times C(7, 3)\) 2. Choosing exactly 3 red marbles and 2 marbles from the other colors: - Choose 3 red marbles from 3: \(C(3, 3)\) - Choose 2 marbles from the remaining 7 of other colors: \(C(7, 2)\) - Total combinations for this scenario: \(C(3, 3) \times C(7, 2)\) 3. Choosing all 5 red marbles is not possible since there are only 3 red marbles in the bag.

Now, we will calculate the total number of combinations for the two valid scenarios. - Total combinations for scenario 1: \(C(3, 2) \times C(7, 3) = \frac{3!}{2!1!} \times \frac{7!}{3!4!} = 3 \times 35 = 105\) - Total combinations for scenario 2: \(C(3, 3) \times C(7, 2) = \frac{3!}{3!0!} \times \frac{7!}{2!5!} = 1 \times 21 = 21\) The total number of sets of five marbles that include at least two red ones is: \(105 + 21 = 126\) So, there are 126 different sets of five marbles with at least two red ones.