What is the probability that 6 (of 26) randomly typed letters are both in alphabetical order and distinct?

I've got a start on it, but can't seem to get the end:

Let $D$ be the event that the 6 letters are unique. Then

$$P(D) = \frac{{26 \choose 6}6!}{26^6}$$

Also where $A$ is the event that the letters are in alphabetical order, we have:

$$P(A) = \frac{{26 \choose 6}}{26^6}$$

Now here is where I'm unsure. If the above is correct, then we're looking for $P(D \cap A) = P(D) + P(A) - P(D \cup A)$. Is this even the right way to go about this, and if so, how do I find the union?


1 Answer 1


The analysis was fine, but once you had the answer you did not see that you had it.

There are $\binom{26}{6}$ ways to choose $6$ distinct letters. For each choice, there is only one way for them to be in alphabetical order.

So your probability is $\dfrac{\binom{26}{6}}{26^6}$. On to the next problem!

  • $\begingroup$ AH! I see now. I was confusing distinct in the set sense with distinct in the permutation sense for what I was calling D. Thanks. $\endgroup$
    – Chester
    Sep 4, 2013 at 23:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .