A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed? Pick one:
A) 1 / 10
B) 1 / 8
C) 1 / 5
D) 1 / 4
E) 1 / 2

Rationale: total number of combinations possible = 10 ! / 2 ! 10 ! / 2 ! total favorable combinations = 9 ! 9 ! - we have considered the two is as one element . since we have two is , they can interchange their positions , without affecting the code . probability = 9 ! / ( 10 ! / 2 ! ) = 9 ! ∗ 2 ! / 10 ! = 1 / 59 ! / ( 10 ! / 2 ! ) = 9 ! ∗ 2 ! / 10 ! = 1 / 5 ans : c
Answer: C