91Ó°ÊÓ

Since we do not include any yellow marbles, we can only choose from the remaining 7 marbles (3 red, 2 green, 1 lavender, and 2 orange). We will use the combination formula to determine the number of ways to choose 5 marbles out of these 7, which is given by: \[C(n, k) = \frac{n!}{k!(n-k)!}\] Here, n = 7 and k = 5.

Using the combination formula, we get: \[C(7, 5) = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2} = 21 \] So, there are 21 ways to choose 5 marbles without any yellow ones.

Now we will have one yellow marble in the set, and we need to find the number of ways to choose the remaining 4 marbles from the other 7 marbles (3 red, 2 green, 1 lavender, and 2 orange). First, we will consider the number of ways to pick one yellow marble out of two, which is: \[C(2, 1) = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = 2 \] Now, we will calculate the number of ways to choose the remaining 4 marbles from the other 7: \[C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35 \] Finally, to find the total number of ways to have one yellow marble and 4 others, we multiply the number of ways to pick one yellow marble times the number of ways to pick the remaining marbles: Sets with one yellow marble = 2 (ways to pick 1 yellow) × 35 (ways to pick 4 other marbles) = 70

Now, we add both cases to get the total number of sets: Total sets = 21 (Case 1: no yellow marbles) + 70 (Case 2: one yellow marble) = 91 There are 91 possible sets of five marbles that include at most one yellow marble.