1. ## Combinatorics

Hello!
I had this exam in January this year and there was a problem I didn't know how to solve, but I know these type of problems are pretty easy so I thought maybe someone here can try and explain it to me.

problem:
In a group of students, 50 are from Sweden, 15 from India, 15 from Pakistan and 10 from Iran.
Four students are selected at random (without replacement).
Calculate the probability that one from each country is selected.

Best regards

2. ## Re: Combinatorics

Originally Posted by ogward
problem:
In a group of students, 50 are from Sweden, 15 from India, 15 from Pakistan and 10 from Iran.
Four students are selected at random (without replacement).
Calculate the probability that one from each country is selected.
There are ninety altogether, how many ways to choose four?
Pick one from each group: $50\cdot 15\cdot 15\cdot 10~.$

3. ## Re: Combinatorics

I don't really get it. You just multiply everything together? then what?

4. ## Re: Combinatorics

Originally Posted by ogward
I don't really get it. You just multiply everything together? then what?
Are you asking us to do this problem for you?

5. ## Re: Combinatorics

I don't know if what I wrote was misleading but I have no idea what so ever how to solve this, that's why I posted it here so maybe someone could explain it to me.

6. ## Re: Combinatorics

Originally Posted by ogward
I don't know if what I wrote was misleading but I have no idea what so ever how to solve this, that's why I posted it here so maybe someone could explain it to me.
Do you understand how probability is measured?
I this case you divide the number of ways to pick one each country by the total number of ways to pick four period.
If you do not understand how to do that you must realize that are we are not tutorial service.
So you must rely on your notes.