How many 3-element subsets are there of set ${1,2,...5n}$ in which number 2 appears? $\binom{3}{5n}$
This is how I would do it if it weren't for the last part (number 2 must appear).
How can I include it?
 A: There is an obvious bijection between three-element subsets of $\{1,2,3,\dots,5n\}$ which contain $2$ and two-element subsets of $\{1,3,4,5,\dots,5n\}$.  No need to break into cases or anything like that.
The answer is $$\binom{5n-1}{2}$$
A: In your question I think you mean ${5n}\choose{3}$ instead of the other way round.
For every 3 element subset of $\{1,2,\ldots, 5n\}$ with $2$ in it think about the same subset without the $2$. For example
$$
\{2, 1, 3\}\rightarrow \{1,3\}\ \\
\{2, 4, 7\}\rightarrow \{4,7\} \\
\{2, 1, 5\} \rightarrow \{1,5\} 
$$
This is a one-to-one correspondence between the subsets of $\{1,2,\ldots, 5n\}$ of size 3 and the subsets of $\{1,3,4,\ldots, 5n\}$ of size two.
The size of the latter family is ${5n-1}\choose{2}$, so this is you answer, namely:
$$
{{5n-1}\choose{2}} = \frac{(5n-1)(5n-2)}{2}.
$$
A: I will assume that the subsets are not ordered, i.e $\{1,2,3\} = \{3,1,2\}$.
So a 3-element subset of $\{1,2,...,5n\}$ is of the form
$$  \{x,y,z\} \mbox{ where } x<y<z \mbox{ and } x,y,z \in \{ 1,2, ... , 5n \} $$
and we will count how many of these exist and contain $2$.
Note that $z$ cannot be $2$, so $0$ ways to do so. If $y=2$ then $x=1$, so there are $$ 5n-2 $$ choices for $z$. If $x=2$, then we have to choose two numbers from $\{ 3,4,...,5n\}$ for $y$ and $z$, $$ {5n-2 \choose 2} $$ ways to do so. Total number of ways
$$ \#ways=0+(5n-2)+ {5n-2 \choose 2} = \frac{(5n-2)(5n-1)}{2} $$
