Questions: Suppose we want to choose 2 colors, without replacement, from the 3 colors red, blue, and green. (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?

Suppose we want to choose 2 colors, without replacement, from the 3 colors red, blue, and green. (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?
Transcript text: Suppose we want to choose 2 colors, without replacement, from the 3 colors red, blue, and green. (a) How many ways can this be done, if the order of the choices is relevant? $\square$ (b) How many ways can this be done, if the order of the choices is not relevant? $\square$
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Solution

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Solution Steps

Step 1: Calculating the Permutation

To find the number of ways to choose 2 items from 3 distinct items, where order matters, we use the permutation formula: \[ P(n, k) = \frac{n!}{(n-k)!} \] Substituting the given values, we get: \[ P(3, 2) = \frac{3!}{(3-2)!} = 6 \]

Step 2: Calculating the Combination

To find the number of ways to choose 2 items from 3 distinct items, where order does not matter, we use the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] Substituting the given values, we get: \[ C(3, 2) = \frac{3!}{2!(3-2)!} = 3 \]

Final Answer:

The number of ways to choose 2 items from 3 distinct items, where order matters, is 6. The number of ways to choose 2 items from 3 distinct items, where order does not matter, is 3.

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