We are given that at a birthday party, there are 3 girls and 3 boys, seated in a row. The boys must take the first and the last seats, and at least two of the girls must not sit next to each other.

Let's break down the solution:

Step 1: Fix the boys' positions

• Boys must take the first and last seats. Therefore, we choose 2 boys to fill the first and last seats. The number of ways to select and arrange these 2 boys in the first and last seats is $3 \times 2 = 6$ ways.

Step 2: Arrange the remaining boy

• The remaining boy has to be seated in one of the four middle seats. There are 4 possible seats for this boy, so there are 4 ways to place the remaining boy.

Step 3: Arranging the girls

• After placing all three boys, there are 3 girls left to place in the remaining 3 seats. However, we need to consider the condition that at least two of the girls should not sit next to each other.

Step 4: Calculate valid arrangements for girls

• We first calculate the total number of ways to arrange the 3 girls in the 3 remaining seats, which is $3! = 6$ ways.

• Now we subtract the cases where all the girls are seated together (i.e., all girls are adjacent). If the girls are seated together, they can be treated as a single "block," leaving 3 "blocks" (2 boys at the ends and the block of girls in the middle) to arrange. There are 2 ways to place the girls' block between the boys, and within that block, the 3 girls can be arranged in $3! = 6$ ways. Hence, the number of ways for all girls to sit together is $2 \times 6 = 12$.

• Therefore, the number of ways in which at least two girls are not seated next to each other is the total number of arrangements of the girls minus the number of arrangements where they are all seated together: [ 6 \times 4 \times (6 - 12) = 24