Assume to start with that within each group the people are indistinguishable.
Place $n$ adults, and then there are $n+1$ gaps into which to place the children. Each group of children is a string of the form b?(gb)*g?
. Denote the number of groups with a left-most boy as $s_b$ and the number of groups with a right-most girl as $s_g$. Then we require $k_1 - s_b = k_2 - s_g$ (the pairs are pairs) and $s_b \le n+1$, $s_g \le n+1$ (we don't have more groups than there are groups).
Given valid $s_b$, $s_g$, we can place the pairs as in the original problem (although since we're considering them indistinguishable, we get $\binom{n+k_1-s_b-1}{k_1-s_b}$ placements). We then place the surplus boys in one of $\binom{n+1}{s_b}$ ways and the surplus girls in $\binom{n+1}{s_g}$ ways.
This leads to the result $\sum_{s_b=0}^{k_1} \sum_{s_g=0}^{k_2} [k_1 - s_b = k_2 - s_g][s_b \le n+1][s_g \le n+1] \binom{n+k_1-s_b-1}{k_1-s_b} \binom{n+1}{s_b} \binom{n+1}{s_g}$ $= \sum_{s_b=0}^{\min(k_1, n+1)} [0 \le s_b + k_2 - k_1 \le \min(k_2, n+1)] \binom{n+k_1-s_b-1}{k_1-s_b} \binom{n+1}{s_b} \binom{n+1}{s_b + k_2 - k_1}$ $= \sum_{s_b=\max(0, k_1-k_2)}^{\min(k_1, n+1, n+1 + k_1 - k_2)} \binom{n+k_1-s_b-1}{k_1-s_b} \binom{n+1}{s_b} \binom{n+1}{s_b + k_2 - k_1}$
Then post-multiply by $n!k_1!k_2!$ to account for the people being distinguishable within their types.