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Luis is packing a bag for vacation. He has 9 unique shirts, but he can only fit 5 in his bag. How many different groups of 5 shirts can he take? ◻

Luis is packing a bag for vacation. He has 99 unique shirts, but he can only fit 55 in his bag.\newlineHow many different groups of 55 shirts can he take?\newline◻

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Full solution

Q. Luis is packing a bag for vacation. He has 99 unique shirts, but he can only fit 55 in his bag.\newlineHow many different groups of 55 shirts can he take?\newline◻
  1. Identify Problem: Luis has 99 unique shirts and wants to choose 55 for his bag. This is a combination problem, not permutation, since the order doesn't matter.
  2. Use Combination Formula: To calculate combinations, we use the formula: C(n,k)=n!k!(n−k)!C(n, k) = \frac{n!}{k!(n-k)!}
  3. Substitute Values: Plug in the values: C(9,5)=9!5!(9−5)!C(9, 5) = \frac{9!}{5!(9-5)!}
  4. Simplify Denominator: Simplify the denominator: C(9,5)=9!(5!4!)C(9, 5) = \frac{9!}{(5!4!)}
  5. Calculate Factorials: Calculate the factorials: 9!=9×8×7×6×5!9! = 9\times8\times7\times6\times5!, 5!=5×4×3×2×15! = 5\times4\times3\times2\times1, and 4!=4×3×2×14! = 4\times3\times2\times1
  6. Cancel Common Factorial: Cancel out the common 5!5! in numerator and denominator: C(9,5)=9×8×7×64×3×2×1C(9, 5) = \frac{9\times8\times7\times6}{4\times3\times2\times1}
  7. Perform Calculation: Perform the calculation: C(9,5)=9×8×7×64×3×2×1=9×2×7×61=126C(9, 5) = \frac{9\times8\times7\times6}{4\times3\times2\times1} = \frac{9\times2\times7\times6}{1} = 126

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