Why Read This Book? - Index of

4.8 Cartesian Products and Cardinality 147
coordinate is f(a2), and so on. Since distinct m-tuples from B m represent distinct
functions from A to B, we have proved the following.
Theorem 4.8.8 If A and B are non-empty sets with |A| = m and |B| = n, then
there are n m distinct functions from A to B.
Here is an informal way to think of Theorem 4.8.8. Let A = Nm and imagine
we have m empty slots numbered 1, 2,...,m. Into each slot we insert precisely
one element of B, where any element of B may be used repeatedly if we like.
Filling the slots in this way is equivalent to defining a function f : Nm → B. There
are n choices of element to place in the first slot. For each of these ways to fill the
first slot, there are n ways to fill the second slot, so that there are n 2 ways to fill
the first two slots. Continuing, we have that there are n m ways to fill all m slots.
Theorem 4.8.8 motivates the notation B A to represent the set of all functions from
A to B.
What if we must fill each of the n slots with one of the m elements of B, but
we are not allowed to use any element of B more than once? How many ways can
this be done? First note that if such a task can be done at all, it must be that m ≤ n.
We have a choice of n objects for the first slot. Then, regardless of which element
of B was placed in the first slot, there are n − 1 possible elements for the second
slot, and so on. Thus the number of ways to fill the m slots with distinct elements
of B is n(n − 1)(n − 2)...(n− m + 1). We write this as
P(n, m) = n(n − 1)(n − 2)...(n− m + 1) =
n!
(n − m)!
(4.23)
which is called the number of permutations of n objects taken m at a time. Filling
the slots with elements of B so that there is no repetition is the same as constructing
a function from B to Nm that is one-to-one. With this, we have an informal proof
of the following. A rigorous proof should be done by induction.
EXERCISE 4.8.9 Let A and B be non-empty sets with |A| = m ≤ n = |B|. Then
there are P(n, m) distinct one-to-one functions from A to B. 18
If Eq. (4.23) is to be taken as meaningful for m = 0, it implies that there is one
way to arrange zero of n objects. It also suggests that there exists a unique (empty)
function from the empty set to any finite set.
If U is a finite univeral set and A is a subset of U, then Exercise 4.6.10 implies
the complement rule, which says
|A| = |U| − � �A C� � (4.24)
Sometimes the task of determining the cardinality of a set is more easily done by
exploiting Eq. (4.24).
18 Induct on m ≥ 1.