Amari has 8 paint colors available, and he wants to mix 3 of them to create a new color. How many different sets of 3 colors can Amari choose?

Identify Problem Type: Identify the type of problem. We need to find the number of combinations of 3 colors that can be chosen from 8 colors. This is a combinatorics problem, specifically a combination problem where order does not matter.Use Combination Formula: Use the combination formula to calculate the number of different sets. The number of ways to choose 3 items from 8 is given by the combination formula: C(n, k) = n! / (k!(n - k)!), where n is the total number of items, k is the number of items to choose, and ＂!＂ denotes factorial.Plug in Values: Plug in the values into the combination formula. Here, n = 8 (total colors) and k = 3 (colors to choose). So, we calculate C(8, 3) = 8! / (3!(8 - 3)!).Simplify Factorials: Simplify the factorial expressions. Calculate 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, 3! = 3 × 2 × 1, and (8 - 3)! = 5! = 5 × 4 × 3 × 2 × 1.Cancel Common Terms: Cancel out common terms in the numerator and the denominator. C(8, 3) = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / ((3 × 2 × 1) × (5 × 4 × 3 × 2 × 1)). The 5 × 4 × 3 × 2 × 1 in the numerator and denominator cancel out, and we are left with: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1).Perform Calculation: Perform the calculation. C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = (8 × 7 × 6) / 6 = 8 × 7 = 56.

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Amari has 8 paint colors available, and he wants to mix 3 of them to create a new color.
How many different sets of 3 colors can Amari choose?

Amari has $8$ paint colors available, and he wants to mix $3$ of them to create a new color. $\newline$ How many different sets of $3$ colors can Amari choose?

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Q. Amari has

8

paint colors available, and he wants to mix

3

of them to create a new color.

\newline

How many different sets of

3

colors can Amari choose?

Identify Problem Type: Identify the type of problem. $\newline$ We need to find the number of combinations of $3$ colors that can be chosen from $8$ colors. This is a combinatorics problem, specifically a combination problem where order does not matter.
Use Combination Formula: Use the combination formula to calculate the number of different sets. $\newline$ The number of ways to choose $3$ items from $8$ is given by the combination formula: $C(n, k) = \frac{n!}{k!(n - k)!}$ , where $n$ is the total number of items, $k$ is the number of items to choose, and “ $!$ ” denotes factorial.
Plug in Values: Plug in the values into the combination formula. $\newline$ Here, $n = 8$ (total colors) and $k = 3$ (colors to choose). So, we calculate $C(8, 3) = \frac{8!}{3!(8 - 3)!}$ .
Simplify Factorials: Simplify the factorial expressions. $\newline$ Calculate $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(8 - 3)! = 5! = 5 \times 4 \times 3 \times 2 \times 1$ .
Cancel Common Terms: Cancel out common terms in the numerator and the denominator. $\newline$ $C(8, 3) = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1)}$ . $\newline$ The $5 \times 4 \times 3 \times 2 \times 1$ in the numerator and denominator cancel out, and we are left with: $\newline$ $C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$ .
Perform Calculation: Perform the calculation. $\newline$ $C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56$ .