# Counting Problem using Permutations

The Question was: In how many ways can the letters of the English alphabet be arranged so that there are exactly 10 letters between a and z?

My approach was the following: In between a and z, there are P(26,10) ways to arrange the 10 letters and then since 16 letters remain, we would also take into account the 16! arrangements of those letters. So, by the multiplication rule, the number of ways is 16!*P(26,10).

There are $24!$ permutations of the letters b through y. For each such permutation, if a is to the left of z, it can appear in any of 15 positions (from before the first letter to before the fifteenth letter), and z appears 10 positions later. Similarly, if z is first, it can appear in any of 15 positions and a appears 10 positions later. So $30\cdot 24!$.

Several problems:

• Using $P(26,10)$ includes arrangements where a and z are not 10 apart. There are 24 letters excluding a and z.

• Once the letters between a and z are chosen, there are $26-(10+2)=14 \neq 16$ remaining letters.

• The number of letters to the left of a. It is possible to have anywhere between $0$ and $26-(10+2)$ letters to the left of a.

• a could come before or after z.

Fixing these gives the correct answer.

Another approach:

• start with $26$ empty cells and put either "a followed by 10 gaps followed by z" or "z followed by 10 gaps followed by a" into it somewhere. How many ways can this be done?

• fill in the remaining cells. How many ways can this be done?