To solve this problem, let's break it down into clear, step-by-step calculations:

Step 1: Arrange the $2$ boys. Since both places require boys, each boy must go to a different place. There are $A_2^2$ ways to arrange the $2$ boys, which is calculated as follows:
$$A_2^2 = 2! = 2$$

Step 2: Arrange the $5$ girls, considering that they must be split between two places. There are two scenarios for dividing the girls:

- Scenario 1: Divide the girls into two groups, with one group having $1$ girl and the other having $4$ girls. The number of ways to select $1$ girl from $5$ is calculated using combinations:
$$C_5^1 = \frac{5!}{1!(5-1)!} = 5$$

- Scenario 2: Divide the girls into two groups, with one group having $2$ girls and the other having $3$ girls. The number of ways to select $2$ girls from $5$ is:
$$C_5^2 = \frac{5!}{2!(5-2)!} = 10$$

Adding the ways from both scenarios for the girls gives us a total of $5 + 10 = 15$ ways to split the girls.

Step 3: Combining the arrangements of boys and girls. Since the boys can be arranged in $2$ ways and the girls can be arranged in $15$ ways (the sum from both scenarios), the total number of ways to distribute the $5$ girls and $2$ boys to two places is:
$$A_2^2 \times (5 + 10) = 2 \times 15 = 30$$

However, we need to account for the fact that there are $2$ places, which has already been implicitly considered in our approach for both boys and girls. Therefore, we don't multiply by 2 again at the end as the initial solution implied; the correct total number of distribution plans is directly the result of our calculation:

$$30$$

Therefore, the correct answer, accounting for the detailed steps and correcting the multiplication error in the final step of the provided solution, is $\boxed{30}$.