Online Course Discussion Forum
MC II-B Combinatorics Self-Paced
Problem 10.3 continued, part b
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 ways can the innkeeper assign the guests to the rooms?
One of the three arrangements was 2,2,1 guests in the 5 rooms.
The video placed the two groups of two first, and got 5C2*3C2*5C2*3C1 ways = 900
I placed the single guest first, and got 5C1*5*4C2*4*3 = 1800 ways.
What did i do wrong?
It is fine to choose the single guest first. There are $\displaystyle \binom{5}{1}\binom{5}{1}$ ways to choose the single room and which of the $5$ friends stays there. Then we need to choose the other two rooms ($\displaystyle \binom{4}{2}$), who is staying in the first room ($\displaystyle \binom{4}{2}$), and who is staying in the second room ($\displaystyle \binom{2}{2}$). Alrogether this gives $$\displaystyle \binom{5}{1}\binom{5}{1}\cdot\binom{4}{2}\binom{4}{2}\binom{2}{2} = 900.$$
Note you had $4\cdot 3$ instead of $\displaystyle \binom{4}{2}\binom{2}{2}$, hence why you got twice as many.
Thanks for the clarification. However, I still have a question.
Why would you choose two rooms (4C2)? Why not just say the first set of two has 4 rooms left, and the second set of two has 3 rooms left? I did it this way because I thought we needed to use permutations as all the rooms and guests were different.
That could work as well. However, since the order in which the pairs of people are selected does not matter, we need to divide by $2!$ to get rid of repetitions. You could have chosen $A$ to be alone, $B$ and $C$ to share the first room $R_1$ (out of $4$ available), and $D$ and $E$ to share the second room $R_2$ (out of $3$ available), or you could have chosen $A$ to be alone, $D$ and $E$ to share the first room $R_2$ (out of $4$ available), and $B$ and $C$ to share the second room $R_1$ (out of $3$ available).
So, choosing two rooms in no particular order, and the two pairs of guests to go in each is the same as choosing the two pairs in no particular order, and then choose the two rooms: $$\binom{4}{2} \cdot \binom{4}{2}\binom{2}{2} = \dfrac{\displaystyle \binom{4}{2}\binom{2}{2}}{2!}\cdot 4 \cdot 3.$$
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