Solution: Number of ways arranging ‘n’ objects out of which ‘r’ objects are identical is n!/ r!

Case I: Let us assume we use two BLACK and 1 RED dot. Thus, the total way of arranging them is 3!/2! = 3

Case II: Similarly, let us assume we use two RED and 1 BLACK dot. Thus, the total way of arranging them is 3!/2! = 3

Case III: All three BLACK: 3!/3! = 1

Case IV: All three RED: 3!/3! = 1

Total number of codes: 3 + 3 + 1 + 1 = 8

The first dot can use RED or BLACK, the second dot also can use RED or BLACK, and the third dot also can do, so 2*2*2=8

Therefore, D is the correct answer.

Answer D