I'm confused as to what you mean by the "same" and "different." Can you rephrase the problem in a new post?
let's call our boxes A,B, and C
and our books 1,2,3,4, and 5
let's do b) first
our books are all the same, so we will just separate them into 3 piles adding to 5
1 1 3
1 2 2
1 3 1
2 1 2
2 2 1
3 1 1
so if our first column is box A, second column is B etc
then there are 6 ways.
d) if the boxes are the same as well, then isn't 1 2 2 the same as 2 1 2?
so there are only 2 distinct ways , namely 1 1 3 and 1 2 2, since the order does not matter.
a) notice in b) we found there are 6 ways to put identical books in 3 distinct boxes.
Now suppose that all the books are different.
Couldn't these 5 books now be arranged in 5! or 24 ways, without changing the count in the 3 distinct boxes?
so with all books different and boxes different, the number of ways would be 6x24 or 144 ways.
That leaves c)
see if you can reason it out.
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