To solve the problem of finding how many sets of five marbles include at most one yellow marble, we can break the task into two parts: sets that contain no yellow marbles and sets that contain exactly one yellow marble.

Step 1: Identify the total number of marbles.

We are given:

• 3 red marbles,
• 1 green marble,
• 1 lavender marble,
• 3 yellow marbles,
• 2 orange marbles.

Thus, the total number of marbles is: $3 \, (\text{red}) + 1 \, (\text{green}) + 1 \, (\text{lavender}) + 3 \, (\text{yellow}) + 2 \, (\text{orange}) = 10 \, \text{marbles}.$

Step 2: Calculate the number of non-yellow marbles.

There are $10 - 3 = 7$ non-yellow marbles (3 red, 1 green, 1 lavender, and 2 orange).

Step 3: Sets with no yellow marbles.

To form a set of 5 marbles without any yellow marbles, we choose all 5 marbles from the 7 non-yellow marbles. The number of ways to do this is given by: $\binom{7}{5} = \frac{7 \times 6}{2 \times 1} = 21.$

Step 4: Sets with exactly one yellow marble.

If we want exactly one yellow marble in the set, we choose 1 yellow marble from the 3 available yellow marbles, and the remaining 4 marbles must be chosen from the 7 non-yellow marbles. The number of ways to choose 1 yellow marble is $\binom{3}{1} = 3$, and the number of ways to choose 4 non-yellow marbles from the 7 non-yellow marbles is $\binom{7}{4}$. We compute $\binom{7}{4}$ as: $\binom{7}{4} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.$ Therefore, the total number of sets with exactly one yellow marble is: $\binom{3}{1} \times \binom{7}{4} = 3 \times 35 = 105.$

Step 5: Total number of sets.

The total number of sets with at most one yellow marble is the sum of the sets with no yellow marbles and the sets with exactly one yellow marble: $21 + 105 = 126.$

Thus, the total number of sets of five marbles that include at most one yellow marble is 126.

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Would you like a detailed breakdown or have any questions? Here are some related questions:

1. How would the calculation change if we wanted exactly two yellow marbles?
2. What if we wanted sets of six marbles instead of five?
3. How does the inclusion-exclusion principle apply to this type of problem?
4. How would the answer change if there were four yellow marbles instead of three?
5. What if the problem asked for no yellow marbles at all?

Tip: When dealing with combinatorics problems, breaking them into cases often simplifies the calculations.