# In how many ways can we arrange the English alphabet so that exactly $10$ of the letters lie between $A$ and $Z$?

In how many ways can we arrange the English alphabet (of 26 letters) so that exactly $$10$$ of them lie between $$A$$ and $$Z$$?

Attempt:

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?

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