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LordHades Group Title

There are 8 electrical engineers and 12 computer engineers who are doing a group project. We need to divide them into 4 groups of 5 so that each group contains exactly 2 electrical engineers and 3 computer engineers. How many ways can this be done?

  • one year ago
  • one year ago

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  1. wio Group Title
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    You want to split up the problem into simple, easily counted tasks.

    • one year ago
  2. wio Group Title
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    I'd first assign each group their electrical engineers, then assign them their computer engineers.

    • one year ago
  3. LordHades Group Title
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    so 8! / 6! + 12! / 9!

    • one year ago
  4. wio Group Title
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    These are sequential tasks, so you multiply their counts...

    • one year ago
  5. Zoodude Group Title
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    8 ways

    • one year ago
  6. LordHades Group Title
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    i don't understand how you got to 8 ways.

    • one year ago
  7. Zoodude Group Title
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    495 is the answer I just checked on the calculator

    • one year ago
  8. LordHades Group Title
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    Yes, but how exactly do you get that?

    • one year ago
  9. wio Group Title
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    For assigning electrical engineers, I would try: \[ \binom{8}{2}\cdot \binom{6}{2}\cdot \binom{4}{2}\cdot \binom{2}{2} \]

    • one year ago
  10. wio Group Title
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    For assigning computer engineers, I would try: \[ \binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3} \]

    • one year ago
  11. wio Group Title
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    The total count would be the product of these two.

    • one year ago
  12. LordHades Group Title
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    Thanks.

    • one year ago
  13. wio Group Title
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    Do you understand it?

    • one year ago
  14. LordHades Group Title
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    Most of it. Why would you multiply rather than add?

    • one year ago
  15. wio Group Title
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    You add when you have an option of doing either task. You multiply when you have two do both tasks.

    • one year ago
  16. LordHades Group Title
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    That would make sense since in this question it's an "and".

    • one year ago
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