Example 14.1

A well known example is Lotto 6/49:

• You may have heard of the following statement:
• “When you play Lotto 6/49, there is a 14 million to 1 chance of hitting the jackpot”, or words to that effect.
• Where did the number 14 million come from?
• We make use of the counting rules of course. B
• y the end of this chapter, you will know where the 14 million come from.

There are essentially 3 counting rules.

This rule can be loosely stated as follows:

If a “process” can be divided into stages, and there are $n_1$ possible outcomes in stage 1, $n_2$ possible outcomes in stage 2, and so on, then the total number of possible outcomes equals

$$n_1\times n_2 \times \cdot$$

Example 14.2

Setup:

• A salesman calls on 2 clients every day.
• Each call can result in a sale (s),. or in a no sale (n).

Question: How many different things can happen at the end of the day?

Instructions:

• We can equate “the process” with his calling on clients,
• stage 1 with calling on the first client with $n_1 = 2$ (s or n),
• stage 2 with calling on the second client with $n_2= 2$ (s or n).
• Total possible outcomes $= 2 \times 2 = 4$.
• Note: ‘S’ is defined as a ‘success’

Solutions: What are these 4 possible outcomes?

• We can see them better if we draw a TREE DIAGRAM.
• Every tree has a root, representing the start of the “process”.
• From this root, we draw the same number of branches as there are possible outcomes in stage 1.
• We then draw a vertical line to represent the end of the first stage.
• This is to be followed by more branches and a vertical line to represent the second stage, and so on.
• The tree diagram representing this first example looks like so.

Going along the different branches, we can see that the four different things are:

• ss, sn, ns, nn.
• ss means sale to the first, sale to the second.
• sn means sale to the first, no sale to the second.
• ns means no sale to the first, sale to the second.
• nn means no sale to both.

In the jargon of Statistics, these 4 things are called Sample Points. Together, they make up a Sample Space (SS):

• We write it like this
• $$SS = \{ss, sn, ns, nn\}$$

Example 14.3

Setup: Suppose this salesman calls on 3 clients every day.

Question: How many different things can happen by the end of the day?

Solution:

• It is easy to be tempted to say the answer is $3 \times 2 = 6$.
• But remember there are now 3 stages to represent the 3 calls, and each call has 2 outcomes.
• The total possible outcomes $= 2 \times 2 \times 2 = 8$.
• The tree diagram is like this

$$SS = \{sss, ssn, sns, snn, nss, nsn ,nns, nnn\}$$

Example 14.4

Setup: A Product Manager is asked to evaluate 2 new products. He is to give it a rating:

• P if he thinks the new product is Profitable in the first year
• N if he thinks it is Not profitable
• B if he thinks it will Breakeven.

Question: In how many ways can he evaluate these 2 new products?

Solutions:

• Again, it is easy to be tempted to say the answer is 3 x 2 = 6.
• But remember there are 2 products, and each can be evaluated in 3 different ways.
• The correct answer is $3\times 3 = 9$.
• And the Sample Space is $$SS = \{PP, PN, PB, NP, NN, NB, BP, BN, BB\}$$

Example 14.5

Setup: BC license plates have the following setup:

• It has 6 characters,
• 3 letters,
• followed by a 3 digit number,
• $$LLL \,   DDD$$

Question: How many different regular licence plates for passenger vehicles can the BC government issue?

Solution: Assuming there are no restrictions,

• There are 26 letters, and 10 (0 to 9) digits.
• So the total $$= 26 \times 26\times 26 \times 10 \times 10\times 10 = 17,576,000 \text{ different plates.}$$

Example 14.6

Setup: How many different postal codes can there be in Canada?

Question: What does a Canadian postal code look like? Do you get the exact same answer as in the example above?

Solution:

• Before we study the second and the third Counting Rules, we have to introduce another Mathematical Notation.
• It is the idea of a FACTORIAL, which is represented by the exclamation mark, i.e. !.
• We write it like this n!, where n is an integer, i.e., a whole number. By definition,
• $$n! = n\cdot(n-1)\cdot (n -2) \cdots (2)\cdot(1)$$

In other words, we start from the number $n$:

• multiply by it 1 less than $n$,
• multiply by 1. less again,
• …until we come to the number 1.

For instance, $$4! = 4\times 3\times 2\times 1 = 24$$

We can also make use of our calculator:

• We see that [x!] is on top of the multiplication sign, $\times$, key.
• Remember anything on top of a key, we have to access it by pressing the 2ND key first.

One more thing to remember, $0! = 1$.

This rule is so important it has a special name. It is called a PERMUTATION. A permutation is an ARRANGEMENT of objects into ORDER. We have emphasized the 2 key words, arrangement and order.

Example 14.7

Setup: There are 3 candidates in a Federal Election:

• a Liberal (LIB),
• a Conservative (CON),
• and an NDP (NDP).

In this example, it should be clear that there is an ORDER:

• One person will get the most number of votes,
• and they become MP.
• A second person will get the second largest number of votes.
• A third person will get the least number of votes.
• There is an ORDER, and it is:

Solution: As we can see from the above listing, there are 6, no more. Let us obtain this same answer, 6, from a mathematical formula. It is

$${}_n P_r = \frac{n!}{(n-r)!}$$

where

• $r\leq n$.
• $n$ is the total number of objects to be arranged. In this example, $n = 3$.
• $P$ simply stands for Permutation. It has NO numeric value.
• $r$ is the number of objects to be arranged into order. In this example, $r = 3$ also.

According to the formula given above, we have

$${}_n P_r = {}_3P_3 = \frac{3!}{(3-3)!}=\frac{3!}{0!}=\frac{6}{1}=6$$

Example 14.8

Setup:

• There are 15 people in a contest (a beauty pageant, blueberry pie in a country fair, whatever)and
• A winner, a first runner-up, and a second runner-up are to be declared.
• Again, in this example, the words winner, first, second suggest ORDER.
• It is, therefore, also a permutation, with $n = 15$ and $r =3$.

$${}_n P_r = {}_{15}P_3 = \frac{15!}{(15-3)!}=\frac{15!}{12!}=\frac{15\times 14 \times 13 \times 12!}{12!}=15\times 14 \times 13 =2730$$

You may find it easier to do by your calculator:

• You will find the nPr key on top of the MINUS sign.
• First, enter n, which is 15,
• Then, press 2ND nPr,
• Then enter r, which is 3.
• Now press the EQUAL sign.
• You should get the same answer of 2,730.

This rule also has a special name. It is called a COMBINATION. A combination is a SELECTION of objects WITHOUT REGARD TO ORDER. Again, we have emphasized the key phrases.

Example 14.9.1

Setup: Five people, call them A, B, C, D, and E for convenience, are running to become vice presidents of an organization. Two are to be chosen.

Question: In how many ways can the president choose 2 from the 5 to become his vice presidents?

Solution:

Let us list them first of all, and then we will give the mathematical formula. The president can choose

• We can see that there are 10.
• The general formula is: $${}_n C_{r}= \binom{n}{r}= \frac{n!}{r!(n-r)!}$$

where

• $r\leq n$.
• $n$ is the total number of objects to be arranged. In this example, $n = 5$.
• $C$ simply stands for Combination. It has NO numeric value.
• $r$ is the number of objects to be arranged into order. In this example, $r = 2$ also.

For our example then,

$${}_n C_{r}={}_5 C_{2} = \binom{5}{2}= \frac{5!}{2!(5-2)!}=\frac{5!}{2!(3)!}=\frac{120}{2\times 6}=10$$

Again, the calculator can do the same thing for us:

• See that nCr is located on top of the PLUS sign.
• Enter 5, 2ND nCr, enter 2, and the EQUAL sign.
• The answer is again 10.

In this example, it should be made clear that the 2 positions are exactly the SAME:

• The 2 people chosen will have the exact SAME title, Vice President (VP):
• In other word, AB is the same as BA, for both A and B will become vice presidents.
• AC is the same as CA, for both A and C will be vice presidents,
• And so on…

Example 14.9.2

Setup: Let us suppose the 2 positions are not the same:

• Suppose the first person to be chosen is to become VP Finance,
• and the second person chosen is to be VP Marketing.
• Then AB is not the same as BA.
• AB would mean that A is to be VP Finance, and B is to be VP Marketing. .
• And BA would mean B is to be VP Finance, and A is to be VP Marketing.

In other words, the ORDER is now important. And this becomes a problem of Permutation. We now have [latex]{}_5 P_{2}= 20 [\latex]. We should expect 20 as our answer, because for every AB, there is a different BA. For every AC, there is a CA, and so on. In summary then, when we select 2 objects out of 5, there are 10 different combinations when the order of selection is NOT important, and there are 20 different permutations when the order of selection IS important. Now that we have considered all three counting rules, let us look at several more examples, so we can be sure which rule applies in the situations described below.

Example 14.10

Setup: A stock broker is asked to rate 5 different stocks. He is to give each of the 5 stocks ratings of A, B, C, or D. Question: In how many ways can he rate these 5 stocks? Solution: Is it Rule 1, or 2, or 3 that applies?  If you have any doubts, go back to Example 3 under Rule 1:

• There, we have a Product Manager, here a stock broker.
• There, we have 2 new products, here 5 stocks.
• There, he can evaluate each of the new products in 3 ways, here in 4 ways.
• It is Rule 1 that applies.

Conclusion: The total number of ways the stock broker can rate these 5 stocks $= 4 \times 4 \times 4\times 4 \times4 = 1,024$

Example 14.11

Setup: An auditor has 12 Income Tax Returns in front of him. He has to select 3 of them at random to audit.

Question: In how many ways can he select 3 from the 12?

Solution: Here, he has to choose 3 from 12. The order is not important and, therefore, a combination: ₁₂C₃ = 220.

Using the calculator:

$${}_{12} C_{3}=\frac{12!}{3!}{9!}=\frac{12\cdot 11\cdot 10\cdot 9!}{9!\cdot 3!}=\frac{12\cdot 11\cdot 10}{6}=220$$

Example 14.12

Setup: In Lotto 6/49, as long as you have the 6 correct numbers, you hit the jackpot.

Question: How many possible tickets can be given out for each draw?

Solution: The order is not important and, therefore, a combination:

$${}_{49} C_{6}=13,983,816$$

Example 14.13

Setup: There are 10 people, 4 accountants and 6 Salespeople. Four of them, 2 accountants and 2 salespeople, are to be chosen to form a committee.

Question: In how many ways can this committee be formed?

Solution: Here we have to choose:

• 2 accountants from 4, and
• 2 salespeople from 6,
• Order does not appear to be important.
• So we have: $${}_4 C_2=6\text{  and  } {}_6 C_{2}=15$$

Now, what do you do to these 2 numbers, 6 and 15?

• Do you add them?
• NO. According to Rule 1, you multiply.
• In the first stage, you choose 2 accountants from 4, and there are 6 ways to do so.
• In the second stage, you choose 2 salespeople from 6, and there are 15 ways to do so.
• All together, there are $6 \times15 = 90$ ways to have this committee formed.