Advanced Mathematical/Combination Question.

[JonDuke]

Question for those very smart at math/combinations.
I tried Googling this but got lost with all the permutation talk and not sure if that's even what I need.

I hope this makes sense. I want to know how many possible combinations could be made with the following scenario:

You need 2 LW and have 30 to choose from
You need 2 RW and have 30 to choose from
You need 2 C and have 30 to choose from
You need 2 D and have 20 to choose from
You need 1 G and have 10 to choose from

Is there a mathematical formula that can tell me how many different combinations are possible?
What is this formula called? So I can see if there's a website that I could go into, enter all the variables, and have it list me all combinations.

[nfotiu]

Originally Posted by JonDuke

Question for those very smart at math/combinations.
I tried Googling this but got lost with all the permutation talk and not sure if that's even what I need.

I hope this makes sense. I want to know how many possible combinations could be made with the following scenario:

You need 2 LW and have 30 to choose from
You need 2 RW and have 30 to choose from
You need 2 C and have 30 to choose from
You need 2 D and have 20 to choose from
You need 1 G and have 10 to choose from

Is there a mathematical formula that can tell me how many different combinations are possible?
What is this formula called? So I can see if there's a website that I could go into, enter all the variables, and have it list me all combinations.

Ignore that, psyang corrected me below.

[psyang]

You want the "choose" function:

a Choose b = a!/[(a-b)!b!]

So, 5 choose 2 = 5!/(3!2!) = 120/(6*2) = 10. So there are 10 ways to choose 2 from a group of 5.

In your case, you want (30c2)*(30c2)*(30c2)*(20c2)*(10c1) = 435*435*435*190*10 = 156 394 462 500 possibilities (156.4 billion)

[JonDuke]

That was quick and easy. Thanks!

[psyang]

Originally Posted by nfotiu

Not sure what it would be called. But pretty sure the right formula would be 30*29*30*29*30*29*20*19*10

You are off by a factor of 8 because you are using the permutation function instead of the choose function. The difference is that permutations can include duplicates since it doesn't take order into account.

So, someone picking player A then player B from a list of 30 players would not be the same as someone picking player B then player A.

Using the choose function, those two are identical, so aren't counted twice. This is why when picking 2 from a group, the choose function is half the permutation function:

30c2 = 435
30p2 = 870

[nfotiu]

Originally Posted by psyang

You want the "choose" function:

a Choose b = a!/[(a-b)!b!]

So, 5 choose 2 = 5!/(3!2!) = 120/(6*2) = 10. So there are 10 ways to choose 2 from a group of 5.

In your case, you want (30c2)*(30c2)*(30c2)*(20c2)*(10c1) = 435*435*435*190*10 = 156 394 462 500 possibilities (156.4 billion)

Yeah, that makes more sense. Ignore my response above, it would assume order mattered.