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Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.

My attempt : $5! ( ^6P_3) =14,400$

Is this correct?

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Yes, your answer is correct.

There are $5$ consonants and $3$ vowels.

Vowels can be selected in $\dbinom{6}{3}$ ways.

The vowels be arranged in $3!$ ways

The consonants can be arranged in $5!$ ways.

In total we have $\dbinom{6}{3}\times3!\times5!=14400$ ways.

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    $\begingroup$ Thank you . I will accept your answer with green tick :) $\endgroup$ – user122343 Aug 23 '18 at 1:21
  • $\begingroup$ @user122343 You are welcome :) $\endgroup$ – Key Flex Aug 23 '18 at 1:22
  • $\begingroup$ Sorry I thought I did. I just did it back. $\endgroup$ – user122343 Aug 25 '18 at 6:01
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The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:

_ C _ C _ C _ C _ C _

You can choose which of those six slots to fill in ${6 \choose 3}$ ways. And for each such choice there are $3!$ ways to put the three vowels.

Thus: $5! \times (6 \cdot 5 \cdot 4)/(3 \cdot 2 \cdot 1) \times 3! = 14400 $ ways.

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  • $\begingroup$ How do you figure 7 slots? There are 6. $\endgroup$ – Paul Childs Aug 23 '18 at 1:27
  • $\begingroup$ You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 \cdot 2 \cdot 1)$ in the denominator. $\endgroup$ – John Lou Aug 23 '18 at 2:41
  • $\begingroup$ @DavidG.Stork $5! \times (6 \cdot 5 \cdot 4)/(3 \cdot 2 \cdot 1) \times 3!=14400$ but not $5! \times (6 \cdot 5 \cdot 4)/(3 \cdot 2 \cdot 1) \times 3!\ne86400$ $\endgroup$ – user572932 Aug 23 '18 at 2:52

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