# Is this the correct number of permutations?

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:

...

# 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)

• Your answer looks correct to me. – Steve Kass Apr 23 '15 at 18:16
• 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 $21$). – Pedro M. Apr 23 '15 at 18:17

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.