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### Subject 10. Principles of Counting

In some cases, it's relatively easy to list and count all possible outcomes. For example, if you flip a dime in the air, there are only two possible outcomes: it will come up with either heads or tails. In other cases, there are a large number of possible outcomes. For example, imagine if you want to choose two stocks from the thousands of issues that trade on the NYSE.

If one thing can be done in n1 ways, and a second thing, given the first, can be done in n2 ways, and so on for k things, then the number of ways the k things can be done is n1 x n2 x n3 ... x nk.

For example, suppose a portfolio manager is making two decisions:

• Which type of instrument to invest in: stocks or bonds?
• Which country to invest in: U.S., Canada, or Germany?

The number of possible ways the manager can make these two decisions is 2 x 3 = 6.

Note that the multiplication rule is applicable if there are two or more groupings. In the preceding example, there are two groups: one for investment instruments and the other for countries. In addition, only one item can be selected from each group.

Suppose that there are n numbers in a group and n slots available. Only one member can be assigned to each slot. The number of ways to assign every number to the n slots is n factorial: n! = n x (n - 1) x (n - 2) x (n - 3) ... x 1. Note that by convention 0! = 1.

For example, five equity analysts are assigned to cover five industries. The number of ways to assign them is 5! = 5 x 4 x 3 x 2 x 1 = 120.

Unlike the multiplication rule, factorial involves only a single group. It involves arranging items within a group, and the order of the arrangement does matter. The arrangement of ABCDE is different from the arrangement of ACBDE.

A combination is a listing in which the order of listing does not matter. This describes the number of ways that we can choose r objects from a total of n objects, where the order in which the r objects is listed does not matter (The combination formula, or the binomial formula):

For example, if you select two of the ten stocks you are analyzing, how many ways can you select the stocks? 10! / [(10 - 2)! x 2!] = 45.

An ordered listing is known as a permutation, and the formula that counts the number of permutations is known as the permutation formula. The number of ways that we can choose r objects from a total of n objects, where the order in which the r objects is listed does matter, is:

For example, if you select two of the ten stocks you are analyzing and invest \$10,000 in one stock and \$20,000 in another stock, how many ways can you select the stocks? Note that the order of your selection is important in this case. 10P2 = 10!/(10 - 2)! = 90

Note that there can never be more combinations than permutations for the same problem, because permutations take into account all possible orderings of items, whereas combinations do not.

#### Practice Question 1

How many license plates are possible if each consists of three letters followed by three digits? Repetition is permitted.

A. 108
B. 12,812,904
C. 17,576,000

26 x 26 x 26 x 10 x 10 x 10 = 17,576,000

#### Practice Question 2

A sandwich shop offers three different breads, five different meats, and four different cheeses. How many sandwich combinations involving exactly one of each bread, meat and cheese option does the shop offer?

3 x 5 x 4 = 60

#### Practice Question 3

A portfolio has 10 stocks. Each stock can be sold or kept in the portfolio. What is the number of possible outcomes?

A. 20
B. 524
C. 1024

The number of outcomes is equal to 210 = 1024.

#### Practice Question 4

A network executive has to decide on the programming line-up for Thursday night. If there are ten shows to choose from, how many six-show line-ups are possible?

A. 60
B. 1,000,000
C. 151,200

10P6 = 10!/(10 - 6)! = 151,200

#### Practice Question 5

When you buy a one-dollar ticket for the Florida Lotto, you are trying to match six numbers drawn from 53 numbers, in any order. How many different tickets are possible?

A. 536
B. 22,957,480
C. 53P6

53C6 = 53!/[(53 - 6)! x 6!] = 22,957,480

#### Practice Question 6

If there are 30 entries in a "best of breed" dog show, in how many ways can 1st, 2nd, and 3rd place be awarded?

A. 4060.
B. 30.
C. 24,360.

30P3 = 30!/(30 - 3)! = 24,360

#### Practice Question 7

The National Honor Society at Milton HS has 30 members. They want to select a committee of 5 members to send to the state convention. How many different ways can they select a committee of 5 members from the total 30 members?

A. 150
B. 14 million
C. 142,506

A committee does not have an order. So, we want the combinations of 30 objects (members of NHS at MHS) taken 5 at a time: 30C5 = 30!/[25! x 5!] = 142506. Thus, there are 142,506 different ways to select a committee of 5 members.

#### Practice Question 8

A company has three different projects that must be assigned to members of its staff. There are 7 staff member being considered for these projects. How many ways are there to assign each project to one individual only?

A. 35
B. 210
C. 840

Use the permutation formula to find how many ways one can combine 3 people out of 7. 7!/4! = 210

#### Practice Question 9

How many ways are there to include 15 different stocks into 3 different portfolios if one portfolio has 3 stocks, another 5, and another portfolio has 7 stocks.

A. 360
B. 360,360
C. 1,575,290

Use the multinomial formula: [15!]/[3! x 5! x 7!] = 360,360.

#### Practice Question 10

A student association is holding elections for President, Vice-President, Treasurer, and Secretary. There are 6 people that have volunteered to be elected for one of these positions. How many ways can these positions be elected from that group of people?

A. 360
B. 520
C. 720

Use the permutation formula: [6!] / (6 -4)! = 360.

#### Practice Question 11

BeigeBox, a computer manufacturer, can construct computer systems with any one of 4 processors, 2 memory configurations, 3 hard drives, 2 monitors, 2 keyboards, and 2 CD drives. How many unique configurations of computer systems can BeigeBox construct?

A. 192
B. 96
C. 384

The multiplication rule of counting states with the number of combinations available when there are n_1 options in one aspect, n_2 in another, and so on, up to n_k, is n_1 * n_2 * ... * n_k. In this case, the number of combinations is 4*2*3*2*2*2 = 192.

#### Practice Question 12

An individual has part of his money tied up in 6 stocks. He decided that each must remain in his portfolio, be sold, or be given to his son. How many possible outcomes does this individual face?

A. 18
B. 64
C. 729

The number of possible outcomes is 36 = 729.

#### Practice Question 13

An individual has two different portfolios that he has been investing in. One portfolio was created for his son and has 5 stocks. The other portfolio he created for himself; it has 7 stocks. He has decided that he will keep, sell, or donate the stocks in his portfolio; he will either keep or sell the stocks in his son's portfolio. How many possible outcomes does this individual face for all the stocks in both portfolios?

A. 4,096
B. 69,984
C. 531,441

The number of possible outcomes is:25(37) = 69,984.

#### Practice Question 14

A coin is tossed five times. What is the probability of obtaining exactly three heads?

A. 3/16
B. 1/4
C. 5/16

Probability of 3 events occurring in 5 trials:

Combination of 3 events in 5 trials * (outcome for each event)x * (outcome for each event)(n-x)
where:
Combination of 3 events in 5 trials = n! / (x! * (x-1)!)
= 5! / (3! * 2!) = 120 / 12 = 10

Probability = 10 * (1/2)3 * (1/2)(5-3) = 5/16

#### Practice Question 15

How many three-digit numbers can you form using the digits 1, 3, 5, 7, and 9 without a repeating a digit within the number to be formed?

A. 20
B. 60
C. 125

#### Practice Question 16

How many three-digit numbers can you form using the digits 1, 3, 5, 7, and 9 if you are allowed to repeat any of the aforementioned digits any number of times within the number to be formed?

A. 20
B. 60
C. 125

#### Practice Question 17

You are reviewing a list of 8 recommended securities and wish to invest in 4. You will put 40% of your capital in one, 30% in another, 20% in the third, and 10% in the last one. How many different ways can you choose among the 8 securities and invest according to your design?

A. 1,980
B. 1,774
C. 1,680

The combination, or binomial formula, gives the number of ways that k objects can be chosen from n items, without regard to the order of choosing. The formula is (n choose k) = n! / [k! *(n-k)!]. However, in this case, the order does matter. If we choose stocks 3, 1, 7, and 5, we don't consider that the same as choosing 5, 7, 1 and 3, because the portfolio weights would differ. We need the general permutation formula, which gives the number of ways that k objects can be chosen from n items, with regard to order. The formula is n_P_k = n! / (n-k)!. Therefore 8! / (8-4)! = 8 * 7 * 6 * 5 = 1,680.

#### Practice Question 18

A portfolio manager has identified three asset classes, with 7, 9, and 6 securities, respectively. He wants to put together a portfolio of three securities, with no more than one security from each asset class. In how many ways can he achieve this?

A. 22
B. 96
C. 378