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anonymous

  • 5 years ago

At a Royal Dynasty Chinese Restaurant, dinner for eight consists of three items from column A, four items from column B, and three items from column C. If columns A, B and C have five, seven, and six items, respectively, how many different dinner combinations are possible?

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  1. anonymous
    • 5 years ago
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    Separate the question into 3 events. First you choose your items from A, then B, then C. The possibilities in A are 5 choose 3=5!/(2!3!)=10. In B we have 7 choose 4=7!/(3!4!)=35. In C we have 6 choose 3=6!/(3!3!)=20. We multiply the possibilities of each event together to find the total, which is 10*35*20=7000 possibilities.

  2. anonymous
    • 5 years ago
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    Thank you!

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