Permutations and combinations

Question

Find the probability that a randomly chosen arrangement of the 9 letters in the word DELIVERED has exactly four letters between the two Ds.

Correct answer: 1/9

A detailed process with explanations would be deeply appreciated!

Answer 1

How many different arrangements are there?

Nine letters into 9 spots gives 9!, but to account for the duplicate letters (3 Es and 2 Ds), divide by 3! 2!.

So the denominator of the probability is 9!/3!2! = 30240.

Of those 30240 how many contain 2 Ds separated by exactly 4 characters?

One possibility is:

D _ _ _ _ D _ _ _

Or:

_D_ _ _ _D _ _

And another:

_ _ D _ _ _ _ D _

And finally:

_ _ _ D_ _ _ _ D

So there are four cases where the two D values are separated by exactly 4 letters. How many different ways can the 7 blank spaces be filled in from the remaining 7 letters (where E is repeated 3 times)?

D 7 6 5 4 D 3 2 1

Looks like 7!. But that has to be divided by 3! to account for the duplicate Es.

7!/3! = 840, but there are 4 cases for each of those 840 possibilities. (4 * 840) = 3360

So the probability is:

4(7!/3!) / 9!/3!2! = 3360 / 30240 = 1/9.