In how many ways can one dozen donuts be selected if the store offers 6 varieties of donuts, and your order includes between 3 and 7 maple donuts?

This problem is an application of the *stars and bars* method with additional restrictions. Here, we are selecting a dozen (12) donuts from 6 different varieties, and the total number of maple donuts must be between 3 and 7.

### Step 1: Define the variables
Let the 6 varieties of donuts be represented as \( x_1, x_2, \dots, x_6 \), where:
- \( x_1 \) represents the number of maple donuts,
- \( x_2, x_3, x_4, x_5, x_6 \) represent the number of the other 5 types of donuts.

We need to find the number of ways to choose the donuts such that the total number of donuts is 12, and the number of maple donuts \( x_1 \) satisfies \( 3 \leq x_1 \leq 7 \).

### Step 2: Break the problem into cases
We will consider the number of maple donuts \( x_1 \) for each case, and then count how many ways we can distribute the remaining donuts among the other varieties.

For each case, the total number of donuts \( x_2 + x_3 + x_4 + x_5 + x_6 \) is \( 12 - x_1 \).

#### Case 1: \( x_1 = 3 \)
If \( x_1 = 3 \), then the remaining 9 donuts need to be distributed among the 5 other varieties. This is a classic stars and bars problem, where the number of ways to distribute 9 indistinguishable donuts into 5 distinguishable varieties is:
\[
\text{Number of ways} = \binom{9+5-1}{5-1} = \binom{13}{4}
\]

#### Case 2: \( x_1 = 4 \)
If \( x_1 = 4 \), then the remaining 8 donuts need to be distributed among the 5 other varieties. The number of ways to distribute 8 donuts is:
\[
\text{Number of ways} = \binom{8+5-1}{5-1} = \binom{12}{4}
\]

#### Case 3: \( x_1 = 5 \)
If \( x_1 = 5 \), then the remaining 7 donuts need to be distributed among the 5 other varieties. The number of ways to distribute 7 donuts is:
\[
\text{Number of ways} = \binom{7+5-1}{5-1} = \binom{11}{4}
\]

#### Case 4: \( x_1 = 6 \)
If \( x_1 = 6 \), then the remaining 6 donuts need to be distributed among the 5 other varieties. The number of ways to distribute 6 donuts is:
\[
\text{Number of ways} = \binom{6+5-1}{5-1} = \binom{10}{4}
\]

#### Case 5: \( x_1 = 7 \)
If \( x_1 = 7 \), then the remaining 5 donuts need to be distributed among the 5 other varieties. The number of ways to distribute 5 donuts is:
\[
\text{Number of ways} = \binom{5+5-1}{5-1} = \binom{9}{4}
\]

### Step 3: Calculate the combinations
Now we calculate the binomial coefficients:

- \( \binom{13}{4} = 715 \)
- \( \binom{12}{4} = 495 \)
- \( \binom{11}{4} = 330 \)
- \( \binom{10}{4} = 210 \)
- \( \binom{9}{4} = 126 \)

### Step 4: Add the results
The total number of ways is the sum of the possibilities for each case:
\[
715 + 495 + 330 + 210 + 126 = 1876
\]

Thus, the total number of ways to select one dozen donuts with between 3 and 7 maple donuts is \( \boxed{1876} \).

Would you like more details or have any questions?

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Here are 5 related questions to consider:
1. How does the stars and bars method work in more detail?
2. What if the number of maple donuts could be any number between 0 and 12?
3. How would the problem change if the store offered more than 6 varieties of donuts?
4. How can generating functions be used to solve problems like this one?
5. What if the selection was made without replacement of donuts?

**Tip**: When working with combinatorics, it's helpful to break problems into manageable cases and apply well-known formulas like stars and bars for counting possibilities efficiently.

Explore how to calculate the number of ways to select one dozen donuts from 6 varieties, with the restriction of including between 3 and 7 maple donuts. This problem uses combinatorics and the stars and bars method to solve, making it ideal for students in Grades 10-12.

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