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?

  • $\begingroup$ 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. $\endgroup$ – lulu Dec 10 '17 at 17:23
  • $\begingroup$ @lulu Please share if you have any other method $\endgroup$ – ami_ba Dec 10 '17 at 17:26
  • $\begingroup$ @lulu You want to say that I could have done it by the famous 'trial and error' method? $\endgroup$ – ami_ba Dec 10 '17 at 17:28
  • $\begingroup$ I want to learn any shorter existing method, so any post will be appreciated $\endgroup$ – ami_ba Dec 10 '17 at 17:30
  • 1
    $\begingroup$ 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) . $\endgroup$ – 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.

  • $\begingroup$ yup! got it😀😀 $\endgroup$ – ami_ba Dec 10 '17 at 18:45

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