The Smith family has 4 sons and 3 daughters. In how many ways can they be seated in a row of 7 chairs such that at least 2 boys are next to each other?
To solve this problem, I will use the principle of inclusion-exclusion, which says that the number of ways to do something is equal to the total number of ways minus the number of ways to not do it. In this case, the total number of ways to seat the 7 children in a row is 7!, since each child can occupy any of the 7 chairs. The number of ways to not have at least 2 boys next to each other is the same as the number of ways to have no boys next to each other. To have no boys next to each other, I need to place the 3 girls in alternate positions, leaving the 4 boys to fill the remaining gaps. There are 3! ways to arrange the girls in alternate positions, and 4! ways to arrange the boys in the gaps. Therefore, the number of ways to have no boys next to each other is 3! * 4!. By the principle of inclusion-exclusion, the number of ways to have at least 2 boys next to each other is 7! - 3! * 4!. Evaluating this expression, I get 7! - 3! * 4! = 5040 - 144 = 4896. 