Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

You want to split up the problem into simple, easily counted tasks.
I'd first assign each group their electrical engineers, then assign them their computer engineers.
so 8! / 6! + 12! / 9!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question