Combinatorics Homework Help: 12 Workers, 4 Jobs, 277200 Solutions

  • Thread starter Thread starter Norway
  • Start date Start date
  • Tags Tags
    Combinatorics
AI Thread Summary
The problem involves assigning 12 workers to four jobs with specific worker requirements: four for the first job, three for the second, three for the third, and two for the last. The correct approach to find the total number of combinations is to use the binomial coefficient for each job assignment and multiply the results. The calculation is expressed as {12 choose 4} multiplied by {8 choose 3}, {5 choose 3}, and {2 choose 2}, leading to the total of 277,200 ways to assign the workers. The discussion highlights the challenge of understanding combinatorial methods for those new to the topic. Overall, the solution emphasizes the importance of applying combinatorial principles to solve the problem effectively.
Norway
Messages
50
Reaction score
3

Homework Statement



An IT-company with 12 hired workers, has been given four jobs. The company wants to use four workers on the first job, three on the second, three on the third and two on the last job. How many ways can the company put these 12 people on the four different jobs?

The answer's supposed to be 277 200.

The Attempt at a Solution



We've never ever done combinatorics like this, and the sole reason I'm doing this is because I'm ahead of the rest of my class. Still, I can't figure out how to figure out this problem. I've tried multiplying in different ways I know from probability math earlier, but with no luck. My guess is that matrixes(?) are needed here, but I've never done this before, so I really hope there are some kind ones around who can help me. :-)

Thanks a lot,
From Norway
 
Physics news on Phys.org
how many ways to pick 2 ppl out of 12 then how many ways to pick 3 from 10 then how many ways to pick 3 from 7, the remaining form the 4... multiply together to get answer
 
So it's something like ...
{12 \choose 4} \cdot {8 \choose 3} \cdot {5 \choose 3} \cdot {2 \choose 2}
?
Thanks anyway :)
 
should get same answer either way
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Replies
39
Views
6K
Replies
11
Views
2K
Replies
3
Views
2K
Replies
1
Views
3K
Replies
108
Views
18K
Replies
5
Views
2K
Replies
11
Views
3K
Back
Top