How many permutations of the English alphabet do NOT have all five vowels appearing consecutively?

What I have:

Since there are $26$ letters in the alphabet and each letter can be used only once, there are $26!$ arrangements of the letters in the alphabet in a string. There are $21$ arrangements of the vowels surrounded by the consonants:

1){string of $5$ vowels}{string of $21$ consonants}

2) {string of $1$ consonant}{string of $5$ vowels}{string of $20$ consonants}


21){string of $21$ consonants}{string of $5$ vowels}

So there are $22(5!)(21!)$ strings of the alphabet in which all $5$ vowels appear consecutively. So there are $26!-22(5!)(21!)$ strings in which the $5$ vowels do not appear consecutively.

(not sure why part of this is showing up in bold font)

  • $\begingroup$ Your answer looks correct to me. $\endgroup$ – Steve Kass Apr 23 '15 at 18:16
  • $\begingroup$ This seems correct to me, although you probably meant to say that there are $22$ arrangements of "all vowels together" surrounded by consonants (instead of $21$). $\endgroup$ – Pedro M. Apr 23 '15 at 18:17

You are correct.

Another way to think of the invalid permutations is to think of the vowels as a single block. Leaving 22 arrangeable elements. $22!$ ways to arrange them. Then, for each arrangement, the vowels can be arranged $5!$ ways.

Giving you $22!5!$ like what you calculated.

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