Answer:
{(A,B),(A,C),(A,D),(A,E),(A,F),(B,C),(B,D),(B,E),(B,F),(C,D),(C,E),(C,F),(D,E),(D,F),(E,F)}
Step-by-step explanation:
We are given the following in the question:
A, B, C, D, E, and F
We have to select 2 letters from these 6 letters such that the order of two letters is not important.
Possible combinations =
[tex]^6C_2 = \displaystyle\frac{6!}{2!(6-2)!} = \frac{6!}{2!\times 4!} = 15[/tex]
Thus, 15 combination are possible.
Combinations are
{(A,B),(A,C),(A,D),(A,E),(A,F),(B,C),(B,D),(B,E),(B,F),(C,D),(C,E),(C,F),(D,E),(D,F),(E,F)}
