To solve this problem, we use combinations and the "stars and bars" theorem, which is helpful for distributing items into groups under certain constraints.

i) Choosing six bagels

Since there are no restrictions on how many of each type of bagel can be chosen, this is a problem of distributing 6 identical items (bagels) into 8 different groups (types of bagels). The solution involves stars and bars:

The formula for distributing $n$ identical items into $k$ groups is:

$\binom{n + k - 1}{k - 1}$

Here, $n = 6$ (bagels to choose), and $k = 8$ (types of bagels). Substituting into the formula:

$\binom{6 + 8 - 1}{8 - 1} = \binom{13}{7}$

Calculate $\binom{13}{7}$:

$\binom{13}{7} = \frac{13!}{7! \cdot (13 - 7)!} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 1716$

Thus, there are 1,716 ways to choose six bagels.

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ii) Choosing a dozen bagels with at least one of each kind

This is a more restrictive problem, as we must ensure at least one of each of the 8 types of bagels is chosen. To simplify, we start by allocating 1 bagel to each type, leaving us with $12 - 8 = 4$ bagels to distribute freely among the 8 types.

Now, we use the stars and bars theorem again to distribute these remaining 4 bagels into the 8 types of bagels:

$\binom{n + k - 1}{k - 1}$

Here, $n = 4$ (remaining bagels), and $k = 8$ (types of bagels). Substituting into the formula:

$\binom{4 + 8 - 1}{8 - 1} = \binom{11}{7}$

Calculate $\binom{11}{7}$:

$\binom{11}{7} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330$

Thus, there are 330 ways to choose a dozen bagels with at least one of each kind.

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Final Answers:

1. 1,716 ways to choose 6 bagels.
2. 330 ways to choose a dozen bagels with at least one of each kind.

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Would you like a detailed explanation for any part? Here are some related questions:

1. How is the "stars and bars" theorem applied in other problems?
2. How do you compute binomial coefficients efficiently?
3. What if the bagel shop offered fewer types of bagels (e.g., 6)?
4. What if the bagel shop placed a maximum limit on how many bagels of each type could be selected?
5. Can this problem be solved using a programming approach?

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