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?


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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.