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(a) A company that makes crayons is trying to decide which 3 colors to include in a promotional mini-box of 3 crayons. The company can choose the 3 mini-box colors from its collection of 70 colors. How many mini-boxes are possible?

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(b) From the 12 albums released by a musician, the recording company wishes to release 8 in a boxed set. How many different boxed sets are possible?

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(a) A company that makes crayons is trying to decide which $3$ colors to include in a promotional mini-box of $3$ crayons. The company can choose the $3$ mini-box colors from its collection of $70$ colors. How many mini-boxes are possible? $\newline$ $\square$ $\newline$ (b) From the $12$ albums released by a musician, the recording company wishes to release $8$ in a boxed set. How many different boxed sets are possible? $\newline$ $\square$

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Solve proportions: word problems

Full solution

Q. (a) A company that makes crayons is trying to decide which

3

colors to include in a promotional mini-box of

3

crayons. The company can choose the

3

mini-box colors from its collection of

70

colors. How many mini-boxes are possible?

\newline

\square

\newline

(b) From the

12

albums released by a musician, the recording company wishes to release

8

in a boxed set. How many different boxed sets are possible?

\newline

\square

Color Combinations Calculation: To find the number of different $3$ -color combinations from $70$ colors, we use the combination formula: $\newline$ $\text{Combinations} = \frac{n!}{r!(n-r)!}$ $\newline$ where $n = 70$ (total colors) and $r = 3$ (colors to choose).
Color Combinations Result: Calculation: $\newline$ $\text{Combinations} = \frac{70!}{3!(70-3)!} = \frac{70 \times 69 \times 68}{3 \times 2 \times 1}$ $\newline$ $\text{Combinations} = 54,340$
Album Combinations Calculation: For the boxed sets of $8$ albums from $12$ , we again use the combination formula: $\newline$ $\text{Combinations} = \frac{n!}{r!(n-r)!}$ $\newline$ where $n = 12$ (total albums) and $r = 8$ (albums to choose).
Album Combinations Result: Calculation: $\newline$ $\text{Combinations} = \frac{12!}{8!(12-8)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$ $\newline$ $\text{Combinations} = 495$

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