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).
On
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?
On
Here's my logic:
There would be 24 letters left after taking A and Z, so the 10 letters can be chosen in C(24,10) ways and then arranged in 10! ways.
Now, having this group of 12 letters leaves us with 14 letters and we can arranged these 15 elements in 15! ways.
Hence the ways that it can be arranged in is: C(24,10) * 10! * 15!
Edit: I missed out that A and Z can be swapped (thanks Jaap) so that would double the number of ways to C(24,10) * 10! * 15! * 2 which is equal to 1.86134521E25
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 $30\cdot 24!$.