Distinguished/Indistinguished Problem

This page is mainly discussing an interesting permutation problem in the back of the book. We have done such problems in our homework, but when I did those problems I always feel confused and may miss out some situations. So that's the reason why I want to write this summary.
I will discuss this problem by giving out some examples and hope this will clarify your confusion.
Problem Type A: Distinguished Boxes vs Indistinguished Items
Example1: If we have 9 distinguished boxes and 6 indistinguished items, how many ways can we arrange the items into the boxes.
This kind of problem is similar to the “fruit basket problem” we discussed in lecture. Treat the indistinguished items as stars(*) and we use bars to separate items into boxes.
So we need 6 stars and 8 bars for this example and therefore,

#ways = C((6+8), 8) = C(14, 8) = 3003.

Problem Type D: Distinguished Boxes vs Distinguished Items
Example4: If we have 3 distinguished boxes and 5 distinguished items, how many ways can we arrange the items into the boxes.
                 Step1: You need to count how many items in each box: e.g. (2 in box1, 2 in box2, 1 in box3) and find all the probability of all kinds of item arrangement.
We can simplify (2 in box1, 2 in box2, 1 in box3) into (2,2,1).
                 Step2: for each fixed combination from step1, calculate the total ways of different arrangement of this kind of fixed combination:
#ways(2,2,1) = C(5,2)*C(3,2) = 15
Step3: Iterate step 1 and 2 for all combinations.
Here we can use a trick to make things a bit easier. We know that #ways from step2 are the same for (2, 1, 2) and (1, 2, 2).

So box arrangement in the form of (0, 1 4), we have P(3, 3) = 3! = 6.
#ways(0,1,4) = C(5,0)*C(5,1)*C(4,4) = 5