<s>[INST] 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?
 [/INST] 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 $$\frac{\binom{4}{2}\bigg(\binom{6}{1}\times\binom{5}{1}\times\binom{4}{2}\times\binom{2}{2}\bigg)}{4^6}=\frac{6 \times 30 \times 6}{4^6}=\frac{72 \times 15}{4^6}$$
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