Four boys and three girls are at a playground.

1. Arrange the boys and identify the number of available spaces for the girls.

First, we arrange the $4$ boys in a straight line. The number of ways to do this is:
$4! = 24$
This arrangement creates $5$ available spaces (between the boys and at the two ends) where the girls can be placed so that no two girls are together:
$\text{\_ B \_ B \_ B \_ B \_}$

2. Arrange the girls into the spaces and calculate the total number of ways.

We then choose $3$ of these $5$ spaces for the $3$ girls to stand in. The number of ways to arrange the girls is:
$^5P_3 = 60$
Using the multiplication principle, the total number of ways the $7$ children can stand such that all girls are separated is:
$24 \times 60 = 1440$

1. Treat the boys as a single block and find the number of ways to arrange this block along with the girls.

By treating the $4$ boys as $1$ single block, we have $3$ girls $+ 1$ block $= 4$ entities to arrange in a line.
$\text{Number of ways to arrange the entities} = 4!$

2. Find the number of ways to arrange the boys within their block, and compute the total number of arrangements.

The $4$ boys can be arranged among themselves within the block in $4!$ ways.
$\text{Total number of ways} = 4! \times 4! = 24 \times 24 = 576$

1. Determine the number of ways for Case 1: 3 boys and 3 girls play.

Case 1: The chosen players are 3 boys and 3 girls.
Number of ways to select the players = $\binom{4}{3} \times \binom{3}{3} = 4 \times 1 = 4$.
To ensure all the girls are separated, we arrange the 3 boys in a circle first.
Number of ways to arrange the boys = $(3-1)! = 2! = 2$.
The boys create 3 distinct spaces between them. The 3 girls must be placed in these spaces.
Number of ways to arrange the girls = $3! = 6$.
Number of ways for Case 1 = $4 \times 2 \times 6 = 48$.

2. Determine the number of ways for Case 2: 4 boys and 2 girls play.

Case 2: The chosen players are 4 boys and 2 girls.
Number of ways to select the players = $\binom{4}{4} \times \binom{3}{2} = 1 \times 3 = 3$.
To ensure the girls are separated, we arrange the 4 boys in a circle first.
Number of ways to arrange the boys = $(4-1)! = 3! = 6$.
The boys create 4 distinct spaces between them. The 2 girls must be placed in 2 of these spaces.
Number of ways to arrange the girls = $^4P_2 = 12$.
Number of ways for Case 2 = $3 \times 6 \times 12 = 216$.

3. Sum the outcomes of both cases to find the total number of ways.

Since these are the only two possible cases to choose 6 players from the 7 children, we sum the results.
Total number of ways = $48 + 216 = 264$.