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?


Yes you are correct. Your answer simplifies:

$${24 \choose 10} \times 15! \times 10! \times 2 = 2\times {24!\over 10! 14!} 15! 10! = 2 \times 15 \times 24!$$

which suggests a different way to count this:

  • Suppose A goes before Z. A can be in any of the first $15$ positions, which then determines Z's position. The other $24$ letters can fill the other positions in any order. This gives $15 \times 24!$ ways.

  • Z goes before A: also $15 \times 24!$ ways.

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