Counting and Probability
Permutations and combinations: Problem type 1

Suppose we want to choose 6 letters, without replacement, from 8 distinct letters.
(a) If the order of the choices is not relevant, how many ways can this be done?
(b) If the order of the choices is relevant, how many ways can this be done?
Explanation
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Step 1 ：Given that we have 8 distinct letters and we want to choose 6 letters from them.

Step 2 ：For part (a), since the order of the choices is not relevant, we can use the combination formula to calculate the number of ways to choose 6 letters from 8. The combination formula is given by \(C(n, k) = \frac{n!}{(n-k)!k!}\), where \(n\) is the total number of items, \(k\) is the number of items to choose, and '!' denotes factorial.

Step 3 ：For part (b), since the order of the choices is relevant, we can use the permutation formula to calculate the number of ways to choose 6 letters from 8. The permutation formula is given by \(P(n, k) = \frac{n!}{(n-k)!}\), where \(n\) is the total number of items, \(k\) is the number of items to choose, and '!' denotes factorial.

Step 4 ：Substituting \(n = 8\) and \(k = 6\) into the combination formula, we get \(C(8, 6) = \frac{8!}{(8-6)!6!} = 28\).

Step 5 ：Substituting \(n = 8\) and \(k = 6\) into the permutation formula, we get \(P(8, 6) = \frac{8!}{(8-6)!} = 20160\).

Step 6 ：Final Answer: (a) The number of ways to choose 6 letters from 8, without replacement and when the order of the choices is not relevant, is \(\boxed{28}\).

Step 7 ：(b) The number of ways to choose 6 letters from 8, without replacement and when the order of the choices is relevant, is \(\boxed{20160}\).