11 From $m$ boys and $n$ girls $(10 \geqslant m>n \geqslant 4)$, 2 people are randomly selected to be team leaders. Let event $A$ denote that the 2 selected people are of the same gender, and event $B$ denote that the 2 selected people are of different genders. If the probability of event $A$ is equal to the probability of event $B$, then the possible values of $(m, n)$ are $\qquad$
(11 From the problem, we have
$$
\begin{array}{l}
P(A)=\frac{\mathrm{C}_{m}^{2}+\mathrm{C}_{n}^{2}}{\mathrm{C}_{m+n}^{2}}, \\
P(B)=\frac{\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}}{\mathrm{C}_{m+n}^{2}},
\end{array}
$$

Since $P(A)=P(B)$, we have
$$
\mathrm{C}_{m}^{2}+\mathrm{C}_{n}^{2}=\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1},
$$

which simplifies to
$$
(m-n)^{2}=m+n .
$$

Thus, $m+n$ is a perfect square, and $9 \leqslant m+n \leqslant 19$, so
$$
\left\{\begin{array} { l } 
{ m + n = 9 , } \\
{ m - n = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
m+n=16, \\
m-n=4,
\end{array}\right.\right.
$$

Solving these, we get
$$
\left\{\begin{array} { l } 
{ m = 6 , } \\
{ n = 3 }
\end{array} \text { (discard) or } \left\{\begin{array}{l}
m=10, \\
n=6 .
\end{array}\right.\right.
$$

Therefore, $(m, n)=(10,6)$.