Question

suppose we want to choose 2 colors, without replacement, from the 3 colors red, blue, and green.

(if necessary, consult a list of formulas.)

(a) how many ways can this be done, if the order of the choices is taken into consideration?

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

Explanation:

Calculate permutations when order matters

Using the Permutations knowledge point

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

$$ P(3, 2) = \frac{3!}{(3-2)!} = \frac{6}{1} = 6 $$

Calculate combinations when order does not matter

Using the Combinations knowledge point

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

$$ C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{6}{2 \times 1} = 3 $$

Answer:

Question a

6

Question b

3