Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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!

share|cite|improve this answer
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. – Chester Sep 4 '13 at 23:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.