Questions: Suppose we want to choose 5 colors, without replacement, from 9 distinct colors. (a) If the order of the choices is taken into consideration, how many ways can this be done? (b) If the order of the choices is not taken into consideration, how many ways can this be done?

Suppose we want to choose 5 colors, without replacement, from 9 distinct colors.
(a) If the order of the choices is taken into consideration, how many ways can this be done?

(b) If the order of the choices is not taken into consideration, how many ways can this be done?

Transcript text: Suppose we want to choose 5 colors, without replacement, from 9 distinct colors. (a) If the order of the choices is taken into consideration, how many ways can this be done? $\square$ (b) If the order of the choices is not taken into consideration, how many ways can this be done? $\square$

Solution

Solution Steps

Step 1: Understanding the Problem

We are given a set of 9 distinct colors and need to choose 5 colors. We have two scenarios to consider:
(a) When the order of selection matters.
(b) When the order of selection does not matter.

Step 2: Calculating Permutations

For part (a), where the order of the choices is taken into consideration, we use permutations. The formula for permutations of choosing $ r $ items from $ n $ items is given by:

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

Here, $ n = 9 $ and $ r = 5 $. Therefore, the number of ways is:

\[ P(9, 5) = \frac{9!}{(9-5)!} = \frac{9 \times 8 \times 7 \times 6 \times 5}{1} = 15120 \]

Step 3: Calculating Combinations

For part (b), where the order of the choices is not taken into consideration, we use combinations. The formula for combinations of choosing $ r $ items from $ n $ items is given by:

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

Here, $ n = 9 $ and $ r = 5 $. Therefore, the number of ways is:

\[ C(9, 5) = \frac{9!}{5! \times (9-5)!} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126 \]