The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is:

Step 1: Case 1: All boys sit together

First, consider the case where all 5 boys sit together. We can treat the group of 5 boys as a single unit. Now we have this unit of boys and 4 girls, making a total of $1+4=5$ entities to arrange. These 5 entities can be arranged in $5!$ ways. Within the group of 5 boys, they can arrange themselves in $5!$ ways. So, the total number of ways for this case is the product of these two arrangements.

Step 2: Case 2: No two boys sit together

Next, consider the case where no two boys sit together. For this to happen, the girls must create spaces for the boys. We arrange the 4 girls first, which can be done in $4!$ ways. This creates 5 possible positions for the boys (before the first girl, between any two girls, and after the last girl). Since no two boys can sit together, each boy must occupy a different position. We need to choose 5 positions out of these 5 available positions and arrange the 5 boys in them. This can be done in $P(5{,}5)$ ways. So, the total number of ways for this case is the product of the arrangements of girls and boys.

Step 3: Total number of ways

Since the two cases (all boys sit together or no two boys sit together) are mutually exclusive, the total number of ways is the sum of the ways calculated for each case.

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