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?
Best Answer
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.