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

42 COUNTING AND COMBINATORICS
Theorem 4.2
If C(n, k) denotes the number of ways in which k objects can be selected
from a set of n distinct objects then
C(n, k) =
P (n, k)
k!
=
n!
k!(n − k)! .
Proof.
Since the number of groups of k elements out of n elements is C(n, k) and
each group can be arranged in k! ways then P (n, k) = k!C(n, k). It follows
that
P (n, k) n!
C(n, k) = =
( n
An alternative notation for C(n, k) is
k
or k > n.
k!
k!(n − k)!
)
. We define C(n, k) = 0 if k < 0
Example 4.7
From a group of 5 women and 7 men, how many different committees consisting
of 2 women and 3 men can be formed What if 2 of the men are
feuding and refuse to serve on the committee together
Solution.
There are C(5, 2)C(7, 3) = 350 possible committees consisting of 2 women
and 3 men. Now, if we suppose that 2 men are feuding and refuse to serve
together then the number of committees that do not include the two men
is C(7, 3) − C(2, 2)C(5, 1) = 30 possible groups. Because there are still
C(5, 2) = 10 possible ways to choose the 2 women, it follows that there are
30 · 10 = 300 possible committees
The next theorem discusses some of the properties of combinations.
Theorem 4.3
Suppose that n and k are whole numbers with 0 ≤ k ≤ n. Then
(a) C(n, 0) = C(n, n) = 1 and C(n, 1) = C(n, n − 1) = n.
(b) Symmetry property: C(n, k) = C(n, n − k).
(c) Pascal’s identity: C(n + 1, k) = C(n, k − 1) + C(n, k).
Proof.
a. From the formula of C(·, ·) we have C(n, 0) = n!
0!(n−0)!
= 1 and C(n, n) =