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Consider all permutations of the twenty-six English letters that start with the letter $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?

I really couldn’t figure out how to do this one, can someone explain the solution

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closed as off-topic by user21820, Gibbs, YiFan, GNUSupporter 8964民主女神 地下教會, darij grinberg Feb 11 at 16:46

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Clearly $x$ must occur after $y$. So the options are

  • $y$ occurs in place $2$ and $x$ occurs in place $4$ or later... $23$ possibilities;
  • $y$ occurs in place $3$ and $x$ occurs in place $6$ or later... $21$ possibilities;
  • and so on;
  • $y$ occurs in place $13$ and $x$ occurs in place $26$... $1$ possibility.

Total, $23+21+\cdots+1=144$. Then arrange the other $23$ letters. Answer $$144\times23!\ .$$

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