Number of Arrangements

Phrophetsam

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Oct 29, 2019
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Jasper Ryanjohn is a big fan of permanent markers. One day, he bought two glass pencil
containers so that he can place 6 of his favorite markers in his desk. These markers are of
six dierent colors - ravishing red, ominous orange, youthful yellow, gorgeous green, bombastic
blue, and velveteen violet.

(a) How many ways can he place the markers in the containers such that each container has
at least one marker?

(b) Suppose that he wants to organize n of his markers. Use inductive reasoning to determine
the number of possible arrangements he can make.
 
Please do as we ask, and show us where you need help. That can mean showing your work up to the point you got stuck, or just asking specific questions. Keep in mind that this site is about giving help, not just giving answers.

If you have no ideas at all, you might start by thinking about how you could place markers in the containers. If you have 6 markers, A, B, C, D, E, and F, and two containers, ______ and ______, what would an arrangement look like? What sort of arrangement is not allowed?

Be sure to tell us what you have learned that might be useful. Do you know how to count combinations, permutations, or subsets, for example?
 
Jasper Ryanjohn is a big fan of permanent markers. One day, he bought two glass pencil containers so that he can place 6 of his favorite markers in his desk. These markers are of six dierent colors - ravishing red, ominous orange, youthful yellow, gorgeous green, bombastic blue, and velveteen violet.
(a) How many ways can he place the markers in the containers such that each container has at least one marker?
(b) Suppose that he wants to organize n of his markers. Use inductive reasoning to determine the number of possible arrangements he can make.
I also want to know the answers to Prof. Peterson's questions but also to these:
It seems correct clear that order has not effect here. Is that correct?
Moreover, each of the containers may contain anywhere from zero to six markers. Is that correct?
If the set \(\displaystyle \|A\|=m\) has \(\displaystyle m\) elements and set \(\displaystyle \|B\|=n\) then there are \(\displaystyle m^n\) functions from \(\displaystyle A\to B\).
That explains why we use the notation \(\displaystyle B^A\) for the set of all functions \(\displaystyle A\to B.\)
But that is a general case, it maybe the not every B is the functional image of some A.
If we require that every B is the image of some A that is a surjection. So \(\displaystyle \|A\|\ge\|B\|\).
The number of surjections(onto functions)\(\displaystyle A\to B\) is if \(\displaystyle m=\|A\|~\&~n=\|B\|\) then
\(\displaystyle \text{sur}j(m,n) = \sum\limits_{j = 0}^n\binom{n}{j} {{{( - 1)}^j}{{(n - j)}^m}} \) is the number of onto functions from \(\displaystyle A\to B\).
For your case \(\displaystyle \|A\|=6 ~\&~\|B\|=2\) but of course that mean that the two class containers are distinguishable.
If not divide by two. I hope that you see how complicated counting questions can be?
 
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