How many words with or without meaning can be formed using all the letters of the word "EQUATION" at a time if vowels and consonants occur together. My answer is $5!3!2!=1440$. Am I right?

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    $\begingroup$ Looks OK to me. There are only 2 groups, 5 distinct vowels and 3 distinct consonants which can be independently permuted and the groups themselves can be permuted. $\endgroup$ – Shailesh Sep 12 '15 at 14:51

In the word "EQUATION" there are 3 consonants (Q,T,N) so there are $3!$ ways to arrange and 5 vowels (E,U,A,I,O) so $5!$ ways to arrange. In whole for both first consonant and then vowels or vice-versa, there are $2!$ ways . so in total $2! \cdot 5! \cdot 3!=1440$ ways.

  • $\begingroup$ Welcome to Math.SE. This Question is almost two years old, and essentially you repeat the confirmation requested by the Question given in a previous Comment, but you are certainly allowed to do so. However I'd encourage you to improve the formatting, both as to capitalization and punctuation/spacing, and learn to post with mathematical expressions using $\LaTeX$. $\endgroup$ – hardmath Jun 15 '17 at 4:57

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