(a) There are $10$ people at a party who form pairs to talk to each other, with everyone talking to someone else (so $5$ pairs in total).
In how many different ways can the $10$ people be talking?(b) Suppose now that there are $11$ people at the party, so that now the guests form $5$ pairs and $1$ person is left by themselves.
How many configurations are possible now?(Original picture of the above text here.)
I've got the idea on how to do (a) where I got $945$ as my answer from $\frac{10!}{(2^5) \cdot 5!}$.
For (b), I solved it by $10 \cdot 8 \cdot 6 \cdot 4 \cdot 2 = 3840$.
Can someone tell me if my answers are right for both (a) and (b).
Thanks
Best Answer
Your answer is correct.
Here is another method that confirms your result.
Line up the $10$ people in some order, say by age. The first person in line can choose a conversation partner in $9$ ways. Remove those two people from the line, which leaves us with eight people in the line. The first person remaining in the line can choose a conversation partner in $7$ ways. Remove those two people from the line, which leaves us with six people in the line. The first person remaining in line can choose a conversation partner in $5$ ways. Remove those two people from the line, which leaves us with four people in the line. The first person remaining in line can choose a conversation partner in $3$ ways. Remove those two people from the line, which leaves us with two people. They must be conversation partners. Hence, the number of possible arrangements is $$9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 945$$ which can be expressed in the form $9!!$. The symbol $!!$ is read double factorial.
Your answer is incorrect since you have not chosen the person who does not have a conversation partner.
We choose which person does not have a conversation partner. That leaves $10$ people who form pairs to talk to each other. Thus, the answer is $11$ times the result above. $$11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 10,395$$