In how many ways can we arrange the English alphabet (of 26 letters) so that exactly $10$ of them lie between $A$ and $Z$?
First we select $10$ letters to put between $A$ and $Z$ in $C(24,10)$.
Now the letters that lie outside get selected automatically.
We consider $[A(10 letters)Z]$ as a single unit and permute this with the rest of alphabets in $15!$ ways.
Letters between $A$ and $Z$ can be permuted in $10!$ ways.
Finally we can also permute $A$ and $Z$ in 2 ways.
So using the rule of product, the required answer would be $C(24,10)*(15!)*(10!)*2$.
Is this correct?