The number of ways to divide 5 people into three groups 
How many ways can 5 people be divided into three teams where each team must have at least one member?

Assumably they can either be put in one group of 3 people then two groups with 1 person, or two groups with 2 people then one group of 1 person.
Hence my answer was $$ ^5C_3 +\, (^5C_2)  \cdot (^3C_2)$$
However the provided answer was 
$$ ^5C_3 +\, (^5C_2)  \cdot (^3C_2) \cdot (1/2)$$
Where did the 1/2 come from?
Thanks.
 A: Two add-ons to the information already given.

  
*
  
*The factor $\frac{1}{2!}$ occurs in fact twice in your example, since we have
  \begin{align*}
&(^5C_3) (^2C_\color{blue}{1})(^1C_{\color{blue}{1}})\color{blue}{\frac{1}{2!}}
+\, (^5C_\color{blue}{2})(^3C_\color{blue}{2})(^1C_1)\color{blue}{\frac{1}{2!}}\\
&\quad=10\cdot2\cdot1\cdot\frac{1}{2}+10\cdot 3\cdot 1\cdot \frac{1}{2}+\\
&\quad=25
\end{align*}
  
*We can reformulate the problem and ask for the number of ways to partition a set consisting of $5$ elements into $3$ non-empty subsets. These numbers are known as Stirling numbers of the second kind ${n\brace k}$.
Here we are looking for
  \begin{align*}
{5\brace 3}=25
\end{align*}

A: When you pick the team as two players, another two, and then just the final one is left over, consider the following: if the five players are named $A, B, C, D, E$, then you might start by picking the duo $\{A, B\}$ and then $\{C, D\}$, which results in the breakdown of $\{\{A, B\}, \{C, D\}, \{E\}\}$ or you might have first picked the duo $\{C, D\}$ and then picked $\{A, B\}$, which would result in the same grouping of $\{\{A, B\}, \{C, D\}, \{E\}\}$.
To avoid this double-counting, you divide in this scenario by $2$. (Or, equivalently, multiply by $1/2$.)
