The 26 letters A, B, ... , Z are arrange in a random order. [Equivalently, the letters are selected sequentially at random without replacement.]
a) What is the probability that A comes before B in the random order?
b) What is the probability that A comes before Z in the random order?
c) What is the probability that A comes just before B in the random order?
Any help would be much appreciated. I was thinking that for part c the answer would be $1/26$ because we have $25!$ ways of having A right before B and $26!$ total arrangements. Not sure how to proceed with a and b, however.
Thank you. For parts (a) and (b) is there a more formal way of getting $1/2$? Such as the formula for the total number of ways we can have $A$ before $B$ over the total number of arrangements? Would it be 25 choose 1 ... 2 choose 1 over $26!$ since we can have a in the first spot and B in any spot after it? Then we can have a in the 2nd spot and B in any spot after it. Also, for part (c), doesn't $A$ have to come immediately before $B$, so wouldn't the probability be $1/26$?