People at a convention are offered 5 tours on each of 3 days. In how many ways can a person arrange to go on a tour planned by this convention ?
I think the correct answer to this question is : ${}_3C_5$
As order does not count.
Is this correct ?
Update :
This is an assignment question and is all the information I've been provided. Let's assume all 6 tours are distinct. A person may go on any number of tours. Does this clarify ?
Yes, order of the tours should matter. I'm assuming that the person must chose one of the five tours each day, and can't repeat tours. You can chose the tours in $\binom53$ ways, which get permuted in $3!$ ways to give the total number of arrangements as $\binom53\cdot3!=P^5_3$
Edit
The case of the person going on any number of tours is a bit more involved. Let us say the number of tours he takes on the first day is $n_1$, on the second day is $n_2$ and the third day is $n_3$. Then, assuming that the person doesn't repeat tours,
$$0\le n_1+n_2+n_3\le5$$
Introduce a dummy variable, say $n$, such that $n_1+n_2+n_3+n=5$ with $0\le n\le5$.
$n_1+n_2+n_3=5-n$. If you're familiar with the Stars and Bars method, you'll know that the number of unique ordered triples $(n_1,n_2,n_3)$ is given by $\frac{(7-n)!}{2!(5-n)!}$. We can select $5-n$ tours out of $5$ in $\binom5{5-n}=\binom5n$ ways, and they can permute in $(5-n)!$ ways (note here even the order of tours in a day matters), giving you a total of $\frac{(7-n)!}{2!(5-n)!}\cdot\binom5n\cdot(5-n)!=\frac{(7-n)!}{2!}\cdot\binom5n$ tours for any value of $n$. The total number of distinct tours is just the summation
$\displaystyle\sum^5_{n=0}\frac{(7-n)!}{2!}\cdot\binom5n=\sum_{n=0}^5\frac{5!(7-n)!}{2!(5-n)!n!}=\sum^5_{n=0}60\cdot\frac{(7-n)(6-n)}{n!}=5056$