combination from two groups?

slapshot136

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Oct 14, 2013
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I have a problem that goes like this:

There are two groups of 5: capitals (A,B,C,D,E) and lowercase (a,b,c,d,e) - how many combinations of 3 can be made by choosing from the two groups if at-least 1 from each group must be present?

I first though:
5 (from either group) * 5 (from the other group) * 8 (sum of whats left over in the two groups),

however I end up with too many due to double counting:

I could choose A,a,B, but that is the same as choosing B,a,A

so I think I need to divide by the number of possibly "orders", but I am not sure how to get that from two groups

any ideas?

Thanks!
 
There are two groups of 5: capitals (A,B,C,D,E) and lowercase (a,b,c,d,e) - how many combinations of 3 can be made by choosing from the two groups if at-least 1 from each group must be present?

You can get around that flaw by choosing two from one set and one from the other.
Either two caps and one lower or other way around.
 
so then it would be 5*4*5 for caps/caps/lower, and then *2 because caps/lower/lower is also valid, which gives me the exact same number as before?
 
so then it would be 5*4*5 for caps/caps/lower, and then *2 because caps/lower/lower is also valid, which gives me the exact same number as before?
No. These are combinations. You have a permutation.
There are \(\displaystyle \dbinom{5}{2}\dbinom{5}{1}\) ways to choose two upper-case and one lower-case. So double that answer.

Here is a second way: find a total, then remove the ones not wanted.
\(\displaystyle \dbinom{10}{3}-\dbinom{5}{3}-\dbinom{5}{3}\)
 
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