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How many words can be formed from the letters of the word "d a u g h t e r" so that the vowels never come together?

There are $3$ vowels and $5$ consonants. I first arranged $5$ consonants in five places in $5!$ ways. $6$ gaps are created. Out of these $6$ gaps, I selected $3$ gaps in ${}_6C_3$ ways and then made the vowels permute in those $3$ selected places in $3!$ ways. This leads me to my answer $5!\cdot {}_6C_3 \cdot 3! = 14400$.

The answer given in my textbook is $36000$. Which cases did I miss? What is wrong in my method?

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marked as duplicate by Namaste, Especially Lime, Martin Sleziak, C. Falcon, Simply Beautiful Art Apr 18 '17 at 20:28

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  • $\begingroup$ The last '6!' should be either 3! or 6. $\endgroup$ – CiaPan Apr 18 '17 at 13:28
  • $\begingroup$ That is a typo. I will edit it. $\endgroup$ – Arishta Apr 18 '17 at 13:29
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I believe you have misinterpreted what the question is asking you. It asks how many ways to arrange the letters so all 3 vowels aren't together, i.e. D$\color{red}{\text{AUE}}$GHTR has all three together. You want to avoid this. The number of ways in which they are all together is $6!\times3!=4320.$ The total number of ways to arrange DAUGHTER is $8!=40320$, so the number of ways to avoid the 3 vowels being together is $40320-4320=36000$.

What is wrong with your method is that you didn't even allow words like T$\color{blue}{\text{AU}}$HRDEG because of the AU, whereas the question allows this to count. Hope that made sense.

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