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.
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