Number of ways of distributing 3 identical physics books and 3 identical math books among 3 kids so that each kid receives at least 1 book is?
My attempt: Say, for physics books, using 'Stars and Bars' method, the number of ways comes out to be 10 (similarly computing for math books); hence total ways given that a kid is allowed to be given 0 books is 100.
Using the inclusion exclusion principle, I subtracted the cases where one kid gets exactly 0 books (which is 48), and added cases where 2 kids get exactly 0 books (which is 3). Doing so, I arrived at 55, but my assignment says the answer is 52.
I went through the following similar question: Distributing 12 books to 3 children. However I cannot seem to spot the mistake.
You are correct. The answer in the text is what you would get if you ignore the possibility that one child receives all the books.
Here is another way to look at the problem. With regard to the physics books, there are three possibilities.
Let's consider these cases in turn.
One child receives all three physics books: There are three ways to select the child who receives all three physics books. Each of the other children must receive a mathematics book. There are three ways to distribute the remaining mathematics book to one of the three children. Hence, there are $3 \cdot 3 = 9$ such distributions.
One child receives two physics book and a different child receives one physics book: There are three ways to pick the child who receives two physics books and two ways to pick the child who receives one physics book. The remaining child must receive a mathematics book. That leaves two mathematics books to distribute to three children. Let $x_k$ denote the number of mathematics books received by the $k$th child. Then we want to determine the number of solutions of the equation $$x_1 + x_2 + x_3 = 2$$ in the nonnegative integers. Such a solution corresponds to placing two addition signs in a row of two ones. For instance, $$1 + + 1$$ corresponds to the solution $x_1 = 1$, $x_2 = 0$, and $x_3 = 1$. The number of such solutions is $$\binom{2 + 2}{2} = \binom{4}{2}$$ since we must choose which two of the four positions required for two ones and two addition signs will be filled with addition signs. Hence, the number of ways the books may be distributed in this case is $$3 \cdot 2 \cdot \binom{4}{2} = 3 \cdot 2 \cdot 6 = 36$$
Each child receives one physics book: Since each child has a book, we are free to distribute the mathematics books in any way we would like. The number of ways we can do this is the number of solutions of the equation $$x_1 + x_2 + x_3 = 3$$ in the nonnegative integers. Such a solution corresponds to placing two addition signs in a row of three ones. There are $$\binom{3 + 2}{2} = \binom{5}{2} = 10$$ possible distributions since we must choose which two of the five positions required for three ones and two addition signs will be filled with addition signs.
Total: Since the three cases are mutually exclusive and exhaustive, the number of ways of distributing the books is $9 + 36 + 10 = 55$, as you found.