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

34 COUNTING AND COMBINATORICS
The different numbers are {11, 12, 13, 21, 22, 23, 31, 32, 33}
Of course, trees are manageable as long as the number of outcomes is not
large. If there are many stages to an experiment and several possibilities at
each stage, the tree diagram associated with the experiment would become
too large to be manageable. For such problems the counting of the outcomes
is simplified by means of algebraic formulas. The commonly used formula is
the Fundamental Principle of Counting which states:
Theorem 3.1
If a choice consists of k steps, of which the first can be made in n 1 ways,
for each of these the second can be made in n 2 ways,· · · , and for each of
these the k th can be made in n k ways, then the whole choice can be made in
n 1 · n 2 · · · · n k ways.
Proof.
In set-theoretic term, we let S i denote the set of outcomes for the i th task,
i = 1, 2, · · · , k. Note that n(S i ) = n i . Then the set of outcomes for the entire
job is the Cartesian product S 1 × S 2 × · · · × S k = {(s 1 , s 2 , · · · , s k ) : s i ∈
S i , 1 ≤ i ≤ k}. Thus, we just need to show that
n(S 1 × S 2 × · · · × S k ) = n(S 1 ) · n(S 2 ) · · · n(S k ).
The proof is by induction on k ≥ 2.
Basis of Induction
This is just Theorem 2.4.
Induction Hypothesis
Suppose
n(S 1 × S 2 × · · · × S k ) = n(S 1 ) · n(S 2 ) · · · n(S k ).
Induction Conclusion
We must show
n(S 1 × S 2 × · · · × S k+1 ) = n(S 1 ) · n(S 2 ) · · · n(S k+1 ).
To see this, note that there is a one-to-one correspondence between the sets
S 1 ×S 2 ×· · ·×S k+1 and (S 1 ×S 2 ×· · · S k )×S k+1 given by f(s 1 , s 2 , · · · , s k , s k+1 ) =