Interpretation of multinomial coefficients divided by factorials in a probability problem I wanted to solve the following problem:

There are 15 people in a party, including Hannah and Sarah. We divide the 15 people into 3 groups, where each group has 5 people. What is the probability that Hannah and Sarah are in the same group?

My solution is as follows:
We want to find the probability of an event A and it has to be equal to:
$P(A)=\frac{|A|}{|S|}$, where $S$ is the sample space
$|S|=\binom{15}{5,5,5}$ i.e. number of ways to arrange 15 people into 3 groups of 5 people
$|A|=3\binom{13}{3,5,5}$ i.e. number of ways of arranging 13 people (15 minus Hannah and Sarah) into 3 groups in such a way that the first group has 2 places "reserved" for Hannah and Sarah. This number must be multiplied by 3 because Hannah and Sarah could as well be both assigned to the second or to the third group.
So: $P(A)=\frac{|A|}{|S|}=\frac{3\binom{13}{3,5,5}}{\binom{15}{5,5,5}}=\frac{2}{7}\approx0.29$
I checked this result in the solution sheet and while the resulting number is the same, the way of achieving it seems different. The solution in the solution sheet looks as follows:
$P(A)=\frac{\frac{1}{2!}\binom{13}{3,5,5}}{\frac{1}{3!}\binom{15}{5,5,5}}\approx0.29$
Firstly, I would like to ask if my way of thinking was correct or did I get the correct result by accident?
Secondly, please can someone explain the solution from the solution sheet to me. I have never seen before a construction with a multinomial divided by a factorial. How to interpret it? Could someone please give me the simplest possible explanation of such a construction, or maybe some other example, or a link to a website that would explain it because unfortunately I can't find anything about it in my notes.
Thank you in advance for your help.
 A: Your solution seems valid to me.
Re the given answer sheet:
The denominator represents the number of ways of choosing;

*

*$5$ out of $15$ for the first group : $~\displaystyle \binom{15}{5}$.

*Then, $5$ out of the remaining $10$ for the second group : $~\displaystyle \binom{10}{5}$.

*Then, $5$ out of the remaining $5$ for the third group : $~\displaystyle \binom{5}{5}$.

In the denominator, the $~\displaystyle \frac{1}{3!}$ represents an overcounting adjustment factor.  Suppose that people $1$ through $5$ went into the 1st group, people $6$ through $10$ went into the 2nd group, and people $11$ through $15$ went into the 3rd group.  The question of whether Hannah and Sarah are in the same group is unaffected by (for example) whether people $6$ through $10$ went into the 1st group, rather than people $1$ through $5$.
The numerator is very similar, except that whichever group has Hannah and Sarah must then choose $3$ out of the remaining $(13).$
So, in the numerator, you have
$~\displaystyle \binom{13}{3} \times \binom{10}{5} \times \binom{5}{5}.$
In the numerator, assume that Hannah and Sarah are people numbers $14$ and $15$.
Consider the grouping where persons $11$ through $13$ went into the Hannah-Sarah group, and people $1$ through $5$ went into another group.
The next group chosen could have been people $6$ through $10$, rather than $1$ through $5$.  You would still end up with the same groups.
Another way of thinking about the overcounting adjustment factors is that in the denominator you have $3$ groups of identical size, while in the numerator, you only have $2$ groups of identical size, since the Hannah-Sarah group is (in effect) a group of $3$ other people.
Personally, I regard the above explanation of the $2$ overcounting adjustment factors as convoluted.  If I was attacking the problem, and I was using your strategy, rather than the strategy at the end of my answer, I would not have employed overcounting adjustment factors.  Instead, I would have reasoned as you did, and assumed that the order that the groups were formed was (consistently) relevant in both the numerator and denominator.

Note that there is a much simpler solution.
Without Loss of Generality, Hannah went into the 1st group.
Then, there are $4$ slots remaining in Hannah's group out of $14$ possible slots, for Sarah to join Hannah.
Thus, the probability is $~\displaystyle \frac{4}{14}.$
