A Probability Course for the Actuaries A Preparation for Exam P/1

40 COUNTING AND COMBINATORICS
Example 4.3
Five different books are on a shelf. In how many different ways could you
arrange them
Solution.
The five books can be arranged in 5 · 4 · 3 · 2 · 1 = 5! = 120 ways
Counting Permutations
We next consider the permutations of a set of objects taken from a larger
set. Suppose we have n items. How many ordered arrangements of k items
can we form from these n items The number of permutations is denoted
by P (n, k). The n refers to the number of different items and the k refers to
the number of them appearing in each arrangement. A formula for P (n, k)
is given next.
Theorem 4.1
For any non-negative integer n and 0 ≤ k ≤ n we have
P (n, k) =
n!
(n − k)! .
Proof.
We can treat a permutation as a decision with k steps. The first step can be
made in n different ways, the second in n − 1 different ways, ..., the k th in
n − k + 1 different ways. Thus, by the Fundamental Principle of Counting
there are n(n − 1) · · · (n − k + 1) k−permutations of n objects. That is,
P (n, k) = n(n − 1) · · · (n − k + 1) = n(n−1)···(n−k+1)(n−k)! = n!
(n−k)! (n−k)!
Example 4.4
How many license plates are there that start with three letters followed by 4
digits (no repetitions)
Solution.
The decision consists of two steps. The first is to select the letters and this
can be done in P (26, 3) ways. The second step is to select the digits and
this can be done in P (10, 4) ways. Thus, by the Fundamental Principle of
Counting there are P (26, 3) · P (10, 4) = 78, 624, 000 license plates
Example 4.5
How many five-digit zip codes can be made where all digits are different
The possible digits are the numbers 0 through 9.