# Arranging the letters of this word so that no two vowels are together

Arrange the letters(every arrangement must contain all letters of the word) of the word 'BENGALI', so that no two vowels are together.

What my cute little brain could find out:
Let me first arrange the vowels...
__ E __ A __ I__
where the underscores contain the consonants. Now, clearly, there will be $^{3}P_3$ arrangements. So my brain tells me to find the ans for the E A I one and then multiply it by $^{3}P_3$
Now my brain thinks for a minute and then says:
"Hey! There are $4$ underscores and how many consonants do you have? Its $4$ Is it not a modified stars and bars problem?"
I thought for a moment, and agreed with my brain. Then it said:
"Find all integer solutions to the equation based on the following conditions:
$x_1+x_2+x_3+x_4=4$, where $x_1,x_4≥0$ and $x_2,x_3≥1$"
And the answer to this is $\binom{2 + 4 - 1}{2} = \binom{2 + 4 - 1}{4 - 1}$(just some honesty!)
"But wait! There are $^{4}P_4$ ways of arranging the consonants. So multiply this by $^{4}P_4$"
And finally by $^{3}P_3$
So my final answer is
$$\binom{2 + 4 - 1}{2}\times^{4}P_4 \times^{3}P_3$$ Am I correct? If yes, is there any better or more efficient way? If yes, would you show that?

• Your method certainly looks correct and I'd say it was optimal. I suppose that someone could point out that using Stars and Bars, or the like, to count the number of $4-$ tuples of non-negative numbers that add to $2$ was overkill. – lulu Dec 10 '17 at 17:23
• @lulu Please share if you have any other method – ami_ba Dec 10 '17 at 17:26
• @lulu You want to say that I could have done it by the famous 'trial and error' method? – ami_ba Dec 10 '17 at 17:28
• I want to learn any shorter existing method, so any post will be appreciated – ami_ba Dec 10 '17 at 17:30
• More or less. There are only two patterns: either both are in one slot ($4$ ways to do it) or they aren't ($\binom 42$ ways to do it) . – lulu Dec 10 '17 at 17:30

There are $4$ consonants, hence $5$ slots to place one of the three vowels. The consonants as well as the vowels can be written in any order. It follows that there are $${5\choose3}\cdot 3!\cdot 4!=1440$$ admissible arrangements of the $7$ letters.