You don't want to mix up the 2 groups for you get a lot of duplicates.
Imagine (1,2,3,4) in one group and (5,6,7,8,9) in the other.
Now when you do 4C3 in the first group, you pick 3 from the 4 say (1,2,3) or ( 1,3,4) etc.
Now when you combine both the groups you can get combinations such as (4,5) (5,6) (2,5) etc.
Finally all we need is set of 5 numbers (1,2,3,4,5) (1,2,4,5,2) is the same. You would get 15 such duplicates.
Ananthakrishnan wrote:I solved it this way.
3 math teachers can be selected in 4 (4C3) ways. Remaining 2 teachers can be selected from the remaining total available 6 teachers in 6C2 = 15 ways. So total number of ways is 4 x 15 = 60 ways.
I know that this answer is wrong. But, where am i going wrong?
