Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

LordHades

  • 3 years ago

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?

  • This Question is Closed
  1. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You want to split up the problem into simple, easily counted tasks.

  2. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'd first assign each group their electrical engineers, then assign them their computer engineers.

  3. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so 8! / 6! + 12! / 9!

  4. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    These are sequential tasks, so you multiply their counts...

  5. Zoodude
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    8 ways

  6. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't understand how you got to 8 ways.

  7. Zoodude
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    495 is the answer I just checked on the calculator

  8. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, but how exactly do you get that?

  9. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For assigning electrical engineers, I would try: \[ \binom{8}{2}\cdot \binom{6}{2}\cdot \binom{4}{2}\cdot \binom{2}{2} \]

  10. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For assigning computer engineers, I would try: \[ \binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3} \]

  11. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The total count would be the product of these two.

  12. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks.

  13. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Do you understand it?

  14. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Most of it. Why would you multiply rather than add?

  15. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You add when you have an option of doing either task. You multiply when you have two do both tasks.

  16. LordHades
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That would make sense since in this question it's an "and".

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy