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How many permutations are there of all the letters of the word PATHFINDER with the following restrictions: 1) no vowels together, 2) all vowels together. I know there are 10! ways to arrange 10 different letters. Not sure how to address the restriction of no vowels together. In case 2, all vowels together, means the three vowels are a group. They can appear at the beginning, the end, or in the 6 spaces in between the 7 consonants. I'm not sure how to set up the calculation....help would be appreciated.

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    $\begingroup$ For "all vowels together," treat the group of three vowels as one [big] letter so that you effectively have 8 letters. Figure out how to order these 8 letters, and then consider how many ways you can order the vowels within the group. $\endgroup$ – angryavian May 30 '19 at 6:00
  • $\begingroup$ Got it! Thanks angrybird : ) $\endgroup$ – Armando W May 30 '19 at 6:10
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For all the vowels together, as pointed out in comments, the AIE can be treated as a "letter" with $3!=6$ variations. Then there are $8!$ permutations of the "letters", yielding $6×8!=241920$ permutations in all.

The calculation when none of the vowels together involves stars and bars, where three vowels need to be placed in the $8$ spaces between and around the consonants. This yields $\binom83$ arrangements of consonants and vowels, and for each arrangement the vowels and consonants can be permuted in $3!×7!$ ways, giving $\binom83×3!×7!=1693440$ valid permutations.

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  • $\begingroup$ Thank you! I see the case for all vowels together clearly. In the case with no vowels together, I see that we're using both combinations and permutations. The use of 8C3 is not quite as clear to me, however, I shall sit with it. Thanks so much for the help! $\endgroup$ – Armando W May 30 '19 at 6:30

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