In how many ways can the letters of ENGRAVER be arranged if vowels need to be separated?

Question

In how many ways can the letters of ENGRAVER be arranged if vowels need to be separated?

There are $6$ spots the AEE can go into to get separated.

And there are $5$ ways to arrange the N,G,R,V,R $\to$ $5!/2!$.

For the AEE way, do I just do $6P3$ or $3!/2!$?

Answer 1

${6\choose3}\frac {5!}{2!}\frac {3!}{2!}$

The first factor is the arrangement of vowels and consonants. The second is the arrangement of consonants, and the 3rd is the arrangement of vowels.

Answer 2

First, we arrange the five consonants, then we insert the vowels.

Arranging the five consonants: Choose two of the five positions for the $R$s, which can be done in $\binom{5}{2}$ ways, then arrange the remaining three distinct consonants in the remaining three positions, which can be done in $3!$ ways. Hence, the consonants can be arranged in $$\binom{5}{2}3!$$ distinguishable ways.

As you can verify, this is equal to your answer $$\frac{5!}{2!}$$ for the number of distinguishable arrangements of the consonants.

This creates six spaces, four between successive consonants and two at the ends of the row. For instance, $$\square R \square N \square G \square R \square V \square$$

Inserting the vowels: To separate the three vowels, we must place them in three of these six spaces. Choose two of the six spaces for the two $E$s and one of the remaining four spaces for the $A$. This can be done in $$\binom{6}{2}\binom{4}{1}$$ distinguishable ways.

Total: Since these choices can be made independently, the number of distinguishable arrangements of the letters of the word $ENGRAVER$ in which no two vowels are adjacent is $$\binom{5}{2}3! \cdot \binom{6}{2}\binom{4}{1}$$

Why are your proposed answers for the number of arrangements of the vowels incorrect?

The quantity $P(6, 3)$ is incorrect since the two $E$s are indistinguishable. Notice that dividing $P(6, 3)$ by the $2!$ ways we could permute the $E$s within an arrangement without producing an arrangement distinguishable from the given arrangement yields $$\frac{P(6, 3)}{2!} = \binom{6}{2}\binom{4}{1}$$ The quantity $\frac{3!}{2!}$ is incorrect because you have not accounted for the $\binom{6}{3}$ ways you could place the two vowels. Notice that $$\binom{6}{3} \cdot \frac{3!}{2!} = \binom{6}{2}\binom{4}{1}$$

Answer 3

Consider ENGRAVIS instead, so the consonants and vowels are distinct.

Now you can place four slots between the consonants and one at either side: $$\rm \_\ C_1\ \_\ C_2\ \_\ C_3\ \_\ C_4\ \_\ C_5\ \_ $$ This can be done in $5!$ ways.

You have to choose three out of the six slots to fill in the vowels, which can be done in $\binom{6}{3}$ ways. Once the choice is made, we can choose among $3!$ ways to fill in the vowels.

Now we take into account that S is actually R and I is actually E, so we have counted each placement four times, precisely $2!\cdot2!$.

Hence the total number is $$ \frac{1}{2!\,2!}\cdot5!\cdot\binom{6}{3}\cdot3! $$