In how many ways you can arrange letters “AAAAFHIKNORRSY” such that the first R is coming right after to the first A

combinatoricspermutations

My problem is:

In how many ways you can arrange letters "AAAAFHIKNORRSY" 
such that the first R is coming right after to the first A?

I know you can solve this problem with;

Think "AR" as one letter. We know that rest of the A and R letters will come after "AR", so first let's choose five spaces. Then put AR to the first space, then permute the other 4 letters, we'll have $\frac{4!}{3!}$ different permutations.

Example:

_ AR _ _ A _ A _ R _ A_ _

Then we can arrange rest of the letters in $8!$ ways.

So the answer is: $\binom{13}{5} * 8! * \frac{4!}{3!}$

What I am curious about is that, my first attempt was using the symmetry of combination.

So I've tried the following:

We're only interested in how "A" and "R" letters' ordered.
So let's permute them in the way we want.

First, we have "AAAARR", let's get an "AR" from here. Then the rest can be arranged in $\frac{4!}{3!} = 4$ ways, this is what we want.

But in total, for all "A" and "R"s, we have $\frac{6!}{4! * 2!} = 15$ orders, therefore $\frac{4}{15}$.

So there are $\frac{14!}{4! * 2!}$ permutations for every letter in total, then, the answer should be $\frac{14!}{4! * 2!} * \frac{4}{15}$.

But unfortunately, it's not. I've solved few problems using this method, and they were true. What am I missing? Is this approach completely wrong?

Best Answer

The book answer is correct. Out of $13$ spots (counting $AR$ as single letter), it places the $8$ distinct digits first. In the remaining five spots, the very first spot is filled by $AR$ and the next four spots are filled in $4$ ways.

The second approach overcounts. It is also counting the words in which first $A$ and first $R$ are separate such as

$$\color{red}{A}FHI\color{red}{R}KNO\color{blue}{AAAR}SY$$

But we want the two letters adjacent.

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