Original Question:
In how many ways can 12 different books be distributed among four children so that (a) each child gets three books? (b) the two oldest children get four books each and the two youngest get two books each?
Solution:
a) $12\choose 3$ * $9\choose 3$ *$6\choose 3$ *$3\choose 3$
b) $12\choose 4$ * $8\choose 4$ *$4\choose 2$ *$4\choose 2$
Personal Logical Questions:
Why do we use rule of products for the solution ? Why not rule of sum ?
My logic is, "If a procedure can be broken down into first and second stages, and if there are m possible outcomes for the first stage, and for each of these outcomes there are n possible outcomes for the second stage, then the total [arrangements] can be [given] by m * n ways" (Professor Kay Wiese). Therefore, $9\choose 3$ is the outcome of $12\choose 3$ and so on ... which makes them related to each other, and we can times them (i.e., Product Rule).
Is my approach to understanding this right ?
Edit:
Logical Question:
When to use the rule of sum?
Consider this question:
Original Question:
A student is to answer seven out of 10 questions on an examination. In how many ways can he make his selection if (a) there are no restrictions? (b) he must answer the first two questions? (c) he must answer at least four of the first six questions?
Solution:
Part c) $6\choose 4$ * $4\choose 3$ + $6\choose 5$ * $4\choose 2$ + $6\choose 6$ * $4\choose 1$
There are n = 10 books, and we can choose r = 7 of the books. If we must choose 4 of the first 6 books, then the remaining three books we need to choose should be from the remaining total books 4. Thus we multiply for example $6\choose 4$ * $4\choose 3$ , because the second stage is the outcome of the first stage, so rule of product. However, we add the possibilities because this set $6\choose 4$ * $4\choose 3$ is independent of this set $6\choose 5$ * $4\choose 2$, and so on.
Am I right ?