How many linear arrangements of $5$ A's , $3$ B's , $2$ C's are there with the first $A$ occurring before the first $B$?

Question

How many linear arrangements of $5$ A's , $3$ B's , $2$ C's are there with the first $A$ occurring before the first $B$?

My attempt

Number of arrangements = $\frac {10!}{5!*3!*2!}$=$2520$

But the answer is $1575$

What is wrong with that?

Please elaborate your help as much as possible

Thanks a lot

Answer 1

An alternate way of thinking of this:

First pick which two of the ten positions in the arrangement are occupied by $C$'s.

Then, the first available remaining position must be an $A$.

Finally, arrange the remaining $A$'s and $B$'s in the remaining positions.

Apply multiplication principle (rule of product) and conclude.

$\binom{10}{2}\cdot \binom{7}{3}$

Answer 2

You should count words starting with $A$, $CA$, and $CCA$, and then add. The first one, for example, is $$\frac{9!}{4!3!2!}$$

Answer 3

Ask what the probability is of having the first $A$ before the first $B$. This is the probability of out of the $A$s and $B$s the first letter is an $A$. So you are pulling $A$s and $B$s out of an urn; there are five $A$s and three $B$s. The probability the first is an $A$ is $5/8$. So your answer is $(5/8)\times2520=1575$.