Problem:
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

Solution:
To solve the problem, we need to arrange the boys and girls in alternate positions. Let's first arrange the boys and girls separately in the row.

- Arranging the boys: There are 4 boys, and they can be arranged in 4! ways.
- Arranging the girls: There are 3 girls, and they can be arranged in 3! ways.

Now, we need to arrange the boys and girls in alternate positions. There are 2 cases to consider:

Case 1: The first person in the row is a boy.
In this case, we have the following arrangement:
B G B G B G B

- The first boy can be any of the 4 boys.
- The second boy can be any of the 3 remaining boys.
- The third boy can be any of the 2 remaining boys.
- The fourth boy can be the only remaining boy.
- The first girl can be any of the 3 girls.
- The second girl can be any of the 2 remaining girls.
- The third girl can be the only remaining girl.

Therefore, the total number of arrangements in this case is 4! x 3! = 144.

Case 2: The first person in the row is a girl.
In this case, we have the following arrangement:
G B G B G B B

- The first girl can be any of the 3 girls.
- The second girl can be any of the 2 remaining girls.
- The third girl can be the only remaining girl.
- The first boy can be any of the 4 boys.
- The second boy can be any of the 3 remaining boys.
- The third boy can be any of the 2 remaining boys.
- The fourth boy can be the only remaining boy.

Therefore, the total number of arrangements in this case is 3! x 4! = 144.

Hence, the total number of arrangements in which the boys and girls are seated in alternate positions is 144 + 144 = 288.

Answer: (a) 144.