Question

Let A = {1, 2, 3, 4} and B = {x, y, z}. (a) List five functions
from AtoB. (b) How many functions f : A→B are there?
(c) How many functions f : A→B are one-to-one? (d) How
many functions g: B →A are there? (e) How many functions
g: B →A are one-to-one? (f) How many functions f : A→B
satisfy f (1) = x? (g) How many functions f : A→B satisfy
f (1) = f (2) = x? (h) How many functions f : A→B satisfy
f (1) = x and f (2) = y?

Answer 1

b)3*3*3*3=81
c)3*2*1=6
d)4*4*4=64
e)4*3*2=24
f)1*3*3*3=27
g)1*1*3*3=9
h)1*1*3*3=9

Answer 2

Very interesting question.

Answer 3

I can help with b)

Answer 4

ok thank you wio

Answer 5

I would find all subsets of size \(k\). There are \(\binom nk\) ways to do this where \(n\) is carnality of \(A\).
They can all be mapped to one of \(3\) elements in \(B\). So \(3^k\) possible mappings, for any such subset.

Answer 6

Doing this for subsets of every size, we get: \[
\sum\binom nk3^k=\sum\binom nk 3^k1^{n-k}=(3+1)^n=4^n
\]

Answer 7

Does that make sense?

Answer 8

\(n=|A|=4\) and so \(4^n=4^4=256\)

Answer 9

It think you are wrong Wio. Each element in A can be assign to three different element in B. So there are 3*3*3*3 such function

Answer 10

You are correct if you must use all elements in A. However I am assuming you can use a subset of the elements of A if you want.

Answer 11

No, in an undegrad course when we say function f:A-> B it is always assumed that the domain is the whole A

Answer 12

Then your answer to \(c\) should be \(0\) since you can't have a one to one function that uses all elements in the domain. You're not even consistent with your own assumptions.

Answer 13

yes, that is right! I was wrong!

Answer 14

Well for me it isn't about right or wrong, but consistency. We can do it your way, but then c is 0. We can do it my way, and c is a bit more of a challenge.

Answer 15

Page 24 (Which is slide 4) covers some of this more clearly.
http://www.cs.toronto.edu/~stacho/macm101-2.pdf

Answer 16

well just ask the student what is the definition of function here

Answer 17

@sexybunch Can you fill us in? Does every element in \(A\) have to be used?

Answer 18

Let's do it @watchmath 's way, but then change c) to be 0

Answer 19

can we edit our answer? I don't know how

Answer 20

let me look it up for you on the definition of function ...if you still need it

Answer 21

yes every element of A can be used wio

Answer 22

must be used you mean?

Answer 23

can be used and must be used are not the same.

Answer 24

yes

Answer 25

can be use

Answer 26

The responses work and I really appreciate your help.