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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?

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Amari has 88 paint colors available, and he wants to mix 33 of them to create a new color.\newlineHow many different sets of 33 colors can Amari choose?\newline \square

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Q. Amari has 88 paint colors available, and he wants to mix 33 of them to create a new color.\newlineHow many different sets of 33 colors can Amari choose?\newline \square
  1. Understand the problem: Understand the problem.\newlineAmari has 88 different paint colors and wants to choose 33 of them to mix. We need to find the number of different combinations of 33 colors he can choose from the 88 available colors.
  2. Recognize order doesn't matter: Recognize that the order in which the colors are chosen does not matter.\newlineThis is a combination problem, not a permutation, because the order of the colors in the mix does not change the outcome.
  3. Use combination formula: Use the combination formula to calculate the number of different sets.\newlineThe number of ways to choose 33 items from 88 is given by the combination formula: C(n,k)=n!k!(nk)!C(n, k) = \frac{n!}{k!(n-k)!}, where nn is the total number of items to choose from, kk is the number of items to choose, "!!" denotes factorial, and C(n,k)C(n, k) is the number of combinations.
  4. Plug in values: Plug in the values into the combination formula.\newlineHere, n=8n = 8 (total colors) and k=3k = 3 (colors to choose). So, C(8,3)=8!3!(83)!=8!3!5!C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}.
  5. Calculate factorials: Calculate the factorials and simplify the expression.\newline8!=8×7×6×5×4×3×2×18! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\newline3!=3×2×13! = 3 \times 2 \times 1\newline5!=5×4×3×2×15! = 5 \times 4 \times 3 \times 2 \times 1\newlineNow, C(8,3)=8×7×6×5×4×3×2×1(3×2×1)×(5×4×3×2×1)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)}
  6. Cancel common terms: Cancel out the common terms in the numerator and the denominator.\newlineC(8,3)=8×7×63×2×1C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}
  7. Perform calculation: Perform the calculation.\newlineC(8,3)=8×7×63×2×1=8×7×66=8×7=56C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56
  8. Conclude with final answer: Conclude with the final answer.\newlineAmari can choose 5656 different sets of 33 colors from the 88 available colors.

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