anonymous
  • anonymous
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?
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
You want to split up the problem into simple, easily counted tasks.
anonymous
  • anonymous
I'd first assign each group their electrical engineers, then assign them their computer engineers.
anonymous
  • anonymous
so 8! / 6! + 12! / 9!

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More answers

anonymous
  • anonymous
These are sequential tasks, so you multiply their counts...
anonymous
  • anonymous
8 ways
anonymous
  • anonymous
i don't understand how you got to 8 ways.
anonymous
  • anonymous
495 is the answer I just checked on the calculator
anonymous
  • anonymous
Yes, but how exactly do you get that?
anonymous
  • anonymous
For assigning electrical engineers, I would try: \[ \binom{8}{2}\cdot \binom{6}{2}\cdot \binom{4}{2}\cdot \binom{2}{2} \]
anonymous
  • anonymous
For assigning computer engineers, I would try: \[ \binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3} \]
anonymous
  • anonymous
The total count would be the product of these two.
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
Do you understand it?
anonymous
  • anonymous
Most of it. Why would you multiply rather than add?
anonymous
  • anonymous
You add when you have an option of doing either task. You multiply when you have two do both tasks.
anonymous
  • anonymous
That would make sense since in this question it's an "and".

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