The word FACETIOUS is the most common word in the English language which has each of the five vowels in order from left to right. How many ways can the letters in the word FACETIOUS be arranged such that all five vowels are in order from left to right?

I am wondering if there is an easier way than bashing and counting all the possiblities.

  • $\begingroup$ Don't just count all possibilities, find them using combinatorics . $\endgroup$ – Anonymous Nov 1 '20 at 15:32
  • $\begingroup$ So we have these 9 letters, thus 9 spaces, 5 of which are occupied by vowels. There are 9 choose 5 ways to insert the vowels, for instance _x_xxxx__x translates to _A_EIOU__S. For each such combination there are 4! ways to order the remaining consonants. $\endgroup$ – LPenguin Nov 1 '20 at 15:37

We start with $9!$ and divide by $5!$ to account for the vowels. This is because there are $9!$ permutations without the extra restriction, and then one in every $5!$ of them have the vowels in the desired order. The answer is $3024.$

  • $\begingroup$ Thank You! It makes sense now! $\endgroup$ – VS Puzzler Nov 1 '20 at 15:40
  • $\begingroup$ @VSPuzzler if this answer was useful to you, don't forget that on this website it is practice to upvote it and click the check mark for the record. $\endgroup$ – Favst Nov 1 '20 at 15:43

There are nine spaces, $5$ vowels that must be in order, and $4$ are consonants. You can pick any letter first and choose where to place it so lets choose to place the $4$ consonants first.

We can put the $F$ in any of the $9$ places. We can put the $C$ in any of the remaining $8$. We can place the $T$ and the $S$ in any of the remaining $7$ and $6$ places. That's $9\times 8\times 7 \times 6$ ways to place the consonants.

The remaining $5$ spots must be filled with the vowels in order. There is only one way left to do that.

So there are $9\times 8\times 7 \times 6$ ways.


Alternatively. There are ${9\choose 4}$ ways to pick spots for the consonants, or ${9\choose 5}$ ways to pick spots of the vowels. For any one of the vowel/consonants placements we can arrange-- in the four spots set for the consonants-- the consonants in $4!$ ways. So there are ${9\choose 4}\times 4!$ or ${9\choose 5} \times 4!$ ways.


Or there are $9!$ ways total to do arrange the nine letters. However the vowels can be ordered in $5!$ ways and we can only accept $1$ of the the $5!$ orders. So of all the $9!$ ways to arrange the nine letters we can only accept $1$ out of $5!$ of them. SO there are $\frac{9!}{5!}$ ways.


Method 1: gives us $9\cdot 8 \cdot 7\cdot 6=3024$ ways.

Method 2a: gives us ${9\choose 4}\cdot 4! = \frac {9!}{5!4!}4!= \frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways. Method 2b: and ${9\choose 5} = \frac {9!}{4!5!} = {9\choose 4}$ gives us the same answer.

Method 3: gives us $\frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways.


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