Question

suppose we want to choose 6 colors, without replacement, from 8 distinct colors.

(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?

Explanation:

Calculate combinations when order is not relevant

Using the Combinations knowledge point

$$ \binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8 \times 7}{2 \times 1} = 28 $$

Calculate permutations when order is relevant

Using the Permutations knowledge point

$$ P(8, 6) = \frac{8!}{(8-6)!} = \frac{8!}{2!} = 8 \times 7 \times 6 \times 5 \times 4 \times 3 = 20160 $$

Answer:

Question 1

(a) How many ways can this be done, if the order of the choices is not relevant?
<blank>28</blank>

Question 2

(b) How many ways can this be done, if the order of the choices is relevant?
<blank>20160</blank>