2
$\begingroup$

A fancy bed and breakfast inn has 5 rooms, each with a distinctive color-coded decor. One day 5 friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than 2 friends per room. In how many can the innkeeper assign the guests to the rooms?

The possible answers are: (A) 2100 (B) 2220 (C) 3000 (D) 3120 (E) 3125

$\endgroup$
  • $\begingroup$ For a contest of this type, cases: (i) all singles; (ii) one couple; (iii) two couples. It is easy to miscount the two couple case. $\endgroup$ – André Nicolas Feb 6 '14 at 9:55
3
$\begingroup$

The innkeeper can organize the guests into rooms with the following unordered frequencies: $$1,1,1,1,1, \qquad 1,1,1,2 \qquad \text{or} \qquad 1,2,2.$$

  • Case $\{1,1,1,1,1\}$: there are $5!$ ways to assign the guests.
  • Case $\{1,1,1,2\}$: there's $5 \times 4$ ways to assign the frequencies to rooms, and $\binom{5}{2} 3!$ ways to assign the guests to rooms (once the frequencies have been assigned).
  • Case $\{1,2,2\}$: there's $5 \times \binom{4}{2}$ ways to assign the frequencies to rooms, and $\binom{5}{2}\binom{3}{2}$ ways to assign the guests to rooms (once the frequencies have been assigned).

In total this gives: $$5!+5 \times 4 \times \binom{5}{2}3!+5 \times \binom{4}{2} \times \binom{5}{2}\binom{3}{2}=2220.$$

$\endgroup$
  • $\begingroup$ Beat me to it ! Was about to post the same. $\endgroup$ – lsp Feb 6 '14 at 10:05
  • $\begingroup$ Can you please explain the $\dbinom{5}{2}3!$ in case {$1 1 1 2$[.I counted it in the following process:5 possibilities for the first,4 for the second,3 for the third and only 1for the last 2.This yields the same answer as yours,but I am interested in knowing how you counted it. $\endgroup$ – rah4927 Feb 6 '14 at 10:26
  • $\begingroup$ The way I see it, we place 2 of the 5 people for the 2-person room: $\binom{5}{2}$, and the remaining 3 people can be put in the non-empty rooms in $3!$ ways. (There's probably several equivalent ways of interpreting these numbers.) $\endgroup$ – Rebecca J. Stones Feb 6 '14 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.