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Eight members of a basketball team should stay in a hotel. The hotel has a triple, two doubles and a single. How many ways can be distributed in different rooms ?.

I have in mind the rooms of two people are different, and I do differentiation between them.

I think this could solve it like this: $$ \binom {8}{3}\binom{5}{2}\binom{3}{2}\binom{1}{1} = 1680 $$

this is equivalent to:

$$ \frac{P_8}{P_3P_2P_2P_1}=\frac{8!}{3!2!2!1!}=1680 $$

Now, I argue that two of them are brothers and they always sleep in the same room, how could take that into account?

Blunt
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  • If the two double rooms are to be considered distinct, your analysis is good. If the only thing that matters is who sleeps with whom, divide by $2$. For the brothers, easiest is cases (i) they are in the $3$-room and (ii) they are in a $2$-room. – André Nicolas Aug 28 '15 at 18:36

1 Answers1

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If the brothers are assigned to the three-bed room, there is a choice of 6 to accompany them, so the total is $6\times\binom{5}{2}\times\binom{3}{2}$

If the brothers are assigned to a two-bed room, there are 2 choices,so the total is $2\times\binom{6}{3}\times\binom{3}{2}$

Adding these gives the total

David Quinn
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