Number of arrangement of the word JANUARY such that N is before Y and no two vowels are next to eachother

Question

My attempt to the question is no vowel next to eachother: 4!/2!*(6P3)/2!. How do I find that N is left to Y?

Answer 1

First we need to determine the number of ways to arrange consonants and vowels so that no two vowels are next to each other. That is the number of ways to solve $x_1+x_2+x_3+x_4=4$ with $x_2, x_3 \geq 1$. (Imagine each $+$ is a vowel.) That's the same as the number of ways to solve $x_1+x_2+x_3+x_4=2$, and stars and bars tells us that's $\binom 52=10$.

Once we've chosen a specific arrangement of consonants and vowels, there are $\frac 12 \cdot 4!=12$ ways to arrange the consonants so that N occurs before Y. There are $3$ ways to arrange the vowels (the U can go in any of the $3$ vowel slots). Thus, the total number of acceptable combinations is $10 \cdot 12 \cdot 3 = 360$.

Answer 2

$\underline{\text{Case 1:} ~Y ~\text{is considered a vowel}}$

In this case, since there are $4$ vowels, they can only be placed in positions $1,3,5,$ and $7$.

This can be done in $(4 \times 3)$ ways, since once the $U$ and $Y$ are placed, the position of the $2$ A's is set.

Then there are $3!$ ways of permuting the $3$ consonants.

So, ignoring the constraint about the $N$ preceding the $Y$, there are $\displaystyle (12 \times 6) = 72$ ways of arranging the letters.

Ordinarily, the probability of $N$ preceding $Y$ is $(1/2)$. Here, there is nothing in the (vowels-separated) constraint to affect this probability.

Therefore, in Case 1, the final computation is

$$\frac{72}{2} = 36.$$

An alternative (intuitive) argument is that for every otherwise satisfying sequence where the $N$ precedes the $Y$, the reverse sequence (right to left) is also an otherwise satisfying sequence. However, in the reverse sequence, the $N$ does not precede the $Y$.

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$\underline{\text{Case 2:} ~Y ~\text{is not considered a vowel}}$

Here, the analysis is similar to that given by Stars and Bars, which is also discussed here.

If you consider the $3$ vowels as bars, and the $(7-3)$ remaining positions as stars, then the $3$ vowels divide the $7$ places into $4$ regions.

Since the $3$ vowels can not be together, the $2$nd and $3$rd region must be $\geq 1$, while the $1$st and $4$th region must be $\geq 0$. This is because the $1$st and $4$th region represent the number of positions before the $1$st vowel, or after the $3$rd vowel.

So, you must compute the number of non-negative integer solutions to

$x_1 + x_2 + x_3 + x_4 = 4$, where $x_2, x_3 \geq 1$.

Seting: $y_1 = x_1$,
$y_2 = x_2 - 1$,
$y_3 = x_3 - 1$,
$y_4 = x_4$

this bijects to the number of non-negative integer solutions to

$y_1 + y_2 + y_3 + y_4 = (4 - 2)$.

By Stars and Bars analysis, this computes to

$\displaystyle \binom{2 + [4-1]}{[4-1]} = \binom{5}{3} = 10.$

Once the position of the $3$ vowels, are set, then

There are $3$ choices where to place the $U$. Placing the $U$ sets the position of the $2$ A's.

Also, there are $4!$ ways of permuting the $4$ consonants.

So, again ignoring the $N$ before $Y$ constraint, there are $(10 \times 3 \times 24) = 720$ satisfactory placements.

Similar to Case 1, there is nothing in the [no $2$ vowels together] constraint that will affect whether the $N$ is before the $Y$.

So, in Case $2$, the computation is

$$\frac{720}{2} = 360.$$