slapshot136
New member
- Joined
- Oct 14, 2013
- Messages
- 4
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?
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!