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?
Best Answer
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!