jayanti wrote:a random 10 letter code is to be formed using letters a b c d e f g h i, and i can be used twice. what is the probability that the code that has 2 I's adjacent to one another will be formed

The solution below assumes that i is the only letter that can be used twice.

Total possible arrangements:
Number of ways to arrange the 10 elements a,b,c,d,e,f,g,h,i,i = 10!/2!.

Good arrangements:
In a good arrangement, the two i's are adjacent to each other.
Number of ways to arrange [ii] as a block with the other 8 letters = number of ways to arrange 9 elements = 9!.

P(i's are adjacent) = (good arrangements)/(total possible arrangements) = 9! / (10!/2!) = (2!*9!) /10! = 1/5.

Shouldn't the total arrangement be 10! as they are 10 different letters.

jayanti wrote:Shouldn't the total arrangement be 10! as they are 10 different letters.

Hey,
They are not 10 different. They are 9 different and 'i' is repeated twice.

To find the probability of forming a code with two adjacent I's, we must find the total number of such codes and divide by the total number of possible 10-letter codes.
The total number of possible 10-letter codes is equal to the total number of anagrams that can be formed using the letters ABCDEFGHII, that is 10!/2! (we divide by 2! to account for repetition of the I's).

To find the total number of 10 letter codes with two adjacent I's, we can consider the two I's as ONE LETTER. The reason for this is that for any given code with adjacent I's, wherever one I is positioned, the other one must be positioned immediately next to it. For all intents and purposes, we can think of the 10 letter codes as having 9 letters (I-I is one). There are 9! ways to position 9 letters.

Probability = (# of adjacent I codes) / (# of total possible codes)
= 9! ÷ (10! / 2! ) = ( 9!2! / 10! ) = (9!2! / 10(9!) ) = 1/5
The correct answer is C.