Splitting six friends into two pairs and two singles Six friends agree to meet at the hotel Acropolis in Athens. It happens that there are four hotels with the same name. Each of the six friends picks one hotel at random and goes there.
What is the probability that two friends end up alone and the rest four in pairs?
My solution:
Since we have 6 friends and 4 possible choices for each the sample space consists of $4^6$ possible events.
There are $\binom{6}{2}$ choices for the first pair and $4$ hotel choices. There are $\binom{4}{2}$ choices for the second pair and $3$ hotel choices. There $\binom{2}{1}$ choices for the first single person and $2$ hotel choices. The last person has only one hotel choice. Since we have four groups and we do not care about order, we have to divide our results by $4!$. Putting it all together:
$$
P(A) = \frac{\frac{4\binom{6}{2}3\binom{4}{2}2\binom{2}{1}1}{4!}}{4^6} \approx 0.0439
$$
The author finds
$$
P(A) = \frac{12\binom{6}{2}\binom{4}{2}}{4^6} \approx 0.2637
$$
Who is correct?
 A: For another approach :
We know that $2$ hotels will take pairs and the rest $2$ will take only an individual. So , select which hotels will take pairs by $C(4,2)$.The rest automatically take one person.
Now , determine who will be alone and who will be pairs by $$\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\binom{2}{2}$$
Then , the answer is $$\frac{\binom{4}{2}\bigg(\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\binom{2}{2}\bigg)}{4^6}=\frac{6 \times 30 \times 6}{4^6}=\frac{72 \times 15}{4^6}$$
A: I solved independently, and got the author's answer with a slightly different method:
Let us imagine the setup as a word, where the letters we can choose from are the hotels (4 possible letters), and we have to pick 6 letters to make a word (the position of the letter corresponds to a person).
First we must choose which 2 letters appear twice, We have $4\choose2$ ways of doing this. After that, we simply generate all distinct permutations, which is $\frac{6!}{ 2! \times 2!}$. Multiplying this and dividing it by the total $4^6$ gives us the author's answer of 0.2637.
Let's dissect the errors in your methodology:

*

*First, you multiplied by 4 and not 3 for the second pair.


*Let's now look at the division by $4!$. I understand the motivation, it is because we don't care about the order in which we do the matching. However, we only need to account for the swaps among pairs themselves, and among the singles themselves, so we don't divide by $4!$, only by $2!2!$.That is because <pair1, pair2, single1, single2> can become <pair2, pair1, single1, single2>,  <pair1, pair2, single2, single1>,  <pair2, pair1, single2, single1>, and all these should be counted as the same, but it can never become <single1, pair2, pair1, single2>, because you are counting the pairs first. Therefore, you must divide by 4 to normalize, not by $4!$ because you are over-dividing for some situations that do not occur.
I believe these modifications would give you the author's answer.
