Solution:

Part (i) How many ways are there to choose six bagels?

This problem involves choosing 6 bagels from 8 different types. Each type can be chosen multiple times, and the order does not matter. This is a stars and bars problem, where the formula for the number of solutions is:

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

where:

• $n = 8$ (number of types of bagels),
• $k = 6$ (number of bagels to choose).

$\text{Number of ways} = \binom{8 + 6 - 1}{6} = \binom{13}{6}$

Using the formula for combinations: $\binom{13}{6} = \frac{13!}{6!(13 - 6)!} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 1716$

So, there are 1,716 ways to choose 6 bagels.

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Part (ii) How many ways are there to choose a dozen bagels with at least one of each kind?

Here, we need to choose 12 bagels with at least one of each kind of bagel. Since there are 8 types of bagels, at least one of each means we start by choosing 1 bagel of each type. This uses up $8$ bagels, leaving us to distribute the remaining $12 - 8 = 4$ bagels among the 8 types.

This is again a stars and bars problem, but now we distribute $4$ identical items (remaining bagels) into $8$ categories (types of bagels). The formula is:

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

where:

• $n = 8$ (number of types),
• $k = 4$ (remaining bagels).

$\text{Number of ways} = \binom{8 + 4 - 1}{4} = \binom{11}{4}$

Using the formula for combinations: $\binom{11}{4} = \frac{11!}{4!(11 - 4)!} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330$

So, 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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Do you want more details or have any questions?

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Related Questions:

1. What is the stars and bars method in combinatorics, and how is it applied here?
2. How would the solution change if no bagel type could be repeated?
3. Can we calculate these results using a different method, like recursive reasoning?
4. How would the answer to part (ii) change if we had fewer than 8 types of bagels?
5. How do these combinatorial problems relate to probability?

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Tip:

For "at least one" problems in combinatorics, always start by fulfilling the minimum requirement and then distribute the remainder using stars and bars or other methods.