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 relevant?(b) How many ways can this be done, if the order of the choices is not relevant?
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 relevant?(b) How many ways can this be done, if the order of the choices is not relevant?
Solution
(a) If the order of the choices is relevant, we are dealing with permutations. The formula for permutations is nPr = n! / (n-r)!. Here, n is the total number of options (3 colors), and r is the number of options chosen (2 colors). So, the calculation would be 3P2 = 3! / (3-2)! = 6 / 1 = 6. Therefore, there are 6 ways to choose 2 colors from 3 if the order is relevant.
(b) If the order of the choices is not relevant, we are dealing with combinations. The formula for combinations is nCr = n! / [r!(n-r)!]. Here, n is the total number of options (3 colors), and r is the number of options chosen (2 colors). So, the calculation would be 3C2 = 3! / [2!(3-2)!] = 6 / (2*1) = 3. Therefore, there are 3 ways to choose 2 colors from 3 if the order is not relevant.
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