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?  Choose the correct choice from the following.
A) 1/10	B) 1/8	C) 1/5	D) 1/4	E) 1/2
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:1/5
<result>C</result>
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?