91Ó°ÊÓ

First, let's find the number of sets that include the lavender marble and don't have any yellow marbles. This means we need to choose 4 more marbles out of the remaining 3 red, 2 green, and 2 orange marbles. Using the combinations formula: \[C(n,k) = \frac{n!}{k!(n-k)!}\] where n is the total number of marbles of the allowed colors, and k is the number of marbles we need to choose. In this case, n = 3 + 2 + 2 = 7, and k = 4. Calculating the combinations: \[C(7,4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35\] So, there are 35 sets of five marbles that include the lavender marble and no yellow marbles.

Now let's find the number of sets that include exactly one yellow marble and don't have the lavender marble. This means we need to choose 4 more marbles out of the remaining 3 red, 2 green, and 2 orange marbles, and there are 2 yellow marbles to choose from. First, we'll choose 1 yellow marble from the 2 available: \[C(2,1) = \frac{2!}{1!(2-1)!} = 2\] Then, we need to choose 4 more marbles from the remaining 7 (3 red, 2 green, and 2 orange) without the lavender marble. In this case, n = 3 + 2 + 2 = 7, and k = 4. Calculating the combinations: \[C(7,4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35\] Multiplying the two result, we have: 2 (yellow choices) * 35 (other color choices) = 70 So, there are 70 sets of five marbles including one yellow marble and no lavender marble.

Now, we need to sum the possibilities for Case 1 (lavender and no yellow) and Case 2 (one yellow and no lavender) to get the total number of possible sets. Total Sets = Case 1 sets + Case 2 sets = 35 + 70 = 105 Therefore, there are 105 possible sets of five marbles that include either the lavender one or exactly one yellow one but not both colors.