Five books are put in three boxes, how many ways are there if:
a) The books are different and the boxes are different?
b) The books are the same and the boxes are different?
c) The books are different and the boxes are the same?
d) The books are the same and the boxes are the same?
I'm confused as to what you mean by the "same" and "different." Can you rephrase the problem in a new post?
Interesting question.
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.
To find the number of ways in each scenario, we can use combinatorics and counting principles.
a) The books are different and the boxes are different:
In this case, we need to distribute five different books among three different boxes. The number of ways to do this can be calculated using the concept of permutations. We have five choices for the first book, four choices for the second book, three choices for the third book, two choices for the fourth book, and one choice for the fifth book. So, the total number of ways is the product of these options:
Total ways = 5 * 4 * 3 * 2 * 1 = 120 ways.
b) The books are the same and the boxes are different:
Since the books are the same, it doesn't matter which box each book goes into. We can think of this scenario as distributing identical items into distinct boxes. Again, the number of ways to do this can be calculated using the concept of permutations. In this case, we have only one choice for each book since they are all the same. So, we have:
Total ways = 1 * 1 * 1 * 1 * 1 = 1 way.
c) The books are different and the boxes are the same:
In this scenario, all books are put into a single box. We can think of it as assigning books to boxes, but since all boxes are the same, it doesn't matter which book goes into which box. Therefore, the number of ways can be calculated using the concept of combinations. Using the formula for combinations, we have:
Total ways = C(5, 1) = 5 ways.
d) The books are the same and the boxes are the same:
Since both the books and boxes are the same, the only way to distribute them is if each box contains all the books. So, there is only one way to do this.
Total ways = 1 way.