Thread: Partitions for surjective functions

Let A be the set of all functions f:{1,2,3,4,5}->{1,2,3} and for i=1,2,3 let Ai denote a subset of the functions f:{1,2,3,4,5}->{1,2,3}\i.

i)What is the size of :
1). A,
2).the sizes of its subsets Ai,and
3).Ai∩Aj (i<j) also
4).A1∩A2∩A3.

ii)Find with justification the number of surjective functions f:{1,2,3,4,5}->{1,2,3}.

iii)Explain how each surjective function gives a 3-partiton of f:{1,2,3,4,5}

iv)How many 3-partitons of f:{1,2,3,4,5} are there altogethere?

Answers.
i)1).3^5........2).2^5 each 3).1^5 each.......4).0
ii)We can use stirling numbers of the second kind to find this out S(5,3).
iii)Because the image of the function is made up of 3 elements?
iv)I don't know...I"m still thinking the same as for part ii)

Some notation. If and then .

It can be shown that if is a surjection then the collection is a partition of

So is your #iv correct?

ah...yes...there 3 in total .....I have the partitions properties in my head I can't describe them here because of the LaTex.

Originally Posted by StefanM

ah...yes...there 3 in total .....I have the partitions properties in my head I can't describe them here because of the LaTex.

Well there are more than three.
There are 150 surjections from a set of five to a set of three.
Each of those surjections defines a three-cell partition of the set of five.

If is a surjection then is a partition of