# Arrangements of “dictionary” with no two consecutive vowels

How many 10-letter words can you form by arranging the word DICTIONARY such that no two vowels are beside each other?

What I did is put the vowels and calculate there are $\dfrac{4!}{2!}$ ways in arranging them. The only problem are the consonants. There should be at least one consonant between the four vowels. Need help here.

• Just a comment: are you counting repetitions of ways of arranging the vowels with the I's? It seems to me there are only $3!$ ways of arranging them. – Andrew Whelan Jan 14 '17 at 17:19
• oh sorry , i will edit – TCSHS Jan 14 '17 at 17:23
• Sorry, $\frac{4!}{2!} =12$, not $3!$ – Andrew Whelan Jan 14 '17 at 17:58

First, arrange the six consonants. There are $6!$ ways to do this. There are $7$ possible places to place the $4$ vowels, with no more than one in each place, so you choose $4$ of the $7$ places.

The total number of arrangements with no consecutive vowels is the number of ways to arrange the consonants, multiplied by the number of ways to permute the vowels, multiplied by the number of ways to arrange the vowels or $6!\cdot \dfrac{4!}{2!} \times \large\binom{7}{4}$.

• Why 7C4? not 7P4? – TCSHS Jan 14 '17 at 18:31

There are $\frac{4!}{2} = 12$ ways of arranging the vowels. After that you need to fill the $3$ slots in between with $1$ letter each. There are $6 \times 5 \times 4 = 120$ ways of doing this. You then have $8$ positions left and $3$ letters to choose to put in any of them. There are $8 \times 9 \times 10 = 720$ ways of doing this, since after you insert one you have created a new position to be filled, unlike the last scenario where you have fixed slots to fill.

Answer: $12 \times 120 \times 720 = 1036800.$