Let's address the two parts of the question:

(a) When the order of the choices matters, we are dealing with permutations. The formula for permutations without replacement is:

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

Here, $n = 12$ (the total number of objects) and $r = 6$ (the number of objects we are choosing). Plugging in these values:

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

(b) When the order of the choices does not matter, we are dealing with combinations. The formula for combinations is:

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

Again, $n = 12$ and $r = 6$, so:

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

Now I can compute these values for you:

1. $P(12, 6) = \frac{12!}{6!}$
2. $C(12, 6) = \frac{12!}{6!6!}$

Do you want me to continue with the detailed calculation or would you like any specific details about these steps?

5 Related Questions:

1. What is the difference between combinations and permutations?
2. How does the factorial function affect the calculation in permutations and combinations?
3. Can you explain an example where order matters versus where it does not?
4. How do combinations apply to probability problems?
5. Are there situations where both combinations and permutations are used?

Tip: Always remember to check whether the order of selection matters when deciding between permutations and combinations.