Questions: Suppose we want to choose 6 letters, without replacement, from 8 distinct letters. (a) How many ways can this be done, if the order of the choices is not relevant? (b) How many ways can this be done, if the order of the choices is relevant?

Suppose we want to choose 6 letters, without replacement, from 8 distinct letters. (a) How many ways can this be done, if the order of the choices is not relevant? (b) How many ways can this be done, if the order of the choices is relevant?
Transcript text: Suppose we want to choose 6 letters, without replacement, from 8 distinct letters. (a) How many ways can this be done, if the order of the choices is not relevant? $\square$ (b) How many ways can this be done, if the order of the choices is relevant? $\square$
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Solution

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Solution Steps

Step 1: Identify the problem type

This is a combinations problem where the order of selection does not matter.

Step 2: Apply the combinations formula

To calculate the number of ways to choose 6 items out of 8 distinct items, the formula is: \[C(n, k) = \frac{n!}{k!(n-k)!}\]

Step 3: Substitute the values into the formula

\[C(8, 6) = \frac{8!}{6!(8-6)!}\]

Step 4: Calculate the result

\[C(8, 6) = 28\]

Final Answer:

The number of combinations is 28.

Step 1: Identify the problem type

This is a permutations problem where the order of selection matters.

Step 2: Apply the permutations formula

To calculate the number of ways to choose 6 items out of 8 distinct items, the formula is: \[P(n, k) = \frac{n!}{(n-k)!}\]

Step 3: Substitute the values into the formula

\[P(8, 6) = \frac{8!}{(8-6)!}\]

Step 4: Calculate the result

\[P(8, 6) = 20160\]

Final Answer:

The number of permutations is 20160.

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