Let's break down the problem:

We are selecting 6 objects from 12 distinct objects without replacement, and the task involves two cases:

(a) The order matters:

This situation involves permutations, where the order of the selection is important. The number of ways to choose and arrange 6 objects out of 12 is given by the permutation formula:

$P(n, k) = \frac{n!}{(n-k)!}$

For $n = 12$ and $k = 6$:

$P(12, 6) = \frac{12!}{(12-6)!} = \frac{12!}{6!}$

(b) The order does not matter:

Here, the problem involves combinations, where the order of the selection is not important. The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

For $n = 12$ and $k = 6$:

$C(12, 6) = \frac{12!}{6!6!}$

Now, I will calculate both answers:

(a) Permutations:

$P(12, 6) = \frac{12!}{6!} = \frac{479,001,600}{720} = 665,280$

(b) Combinations:

$C(12, 6) = \frac{12!}{6!6!} = \frac{479,001,600}{720 \times 720} = 924$

So:

• (a) 665,280 ways if order matters.
• (b) 924 ways if order does not matter.

Would you like further details on this solution, or any questions about the steps?

Here are 5 related questions:

1. What is the difference between permutations and combinations?
2. How does the formula change if we were choosing 5 objects instead of 6?
3. What happens if we allow replacement in this problem?
4. How would the problem change if the 12 objects were not distinct?
5. Can you generalize the formula for any values of $n$ and $k$?

Tip: Always remember that permutations are for ordered selections, while combinations are for unordered selections.