**Quantitative Methods: Basic Concepts**

**Reading 8. Probability Concepts**

**Learning Outcome Statements**

o. identify the most appropriate method to solve a particular counting problem and solve counting problems using the factorial, combination, and permutation concepts.

*CFA Curriculum, 2020, Volume 1*

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### Subject 10. Principles of Counting

If one thing can be done in n

_{1}ways, and a second thing, given the first, can be done in n

_{2}ways, and so on for k things, then the number of ways the k things can be done is n

_{1}x n

_{2}x n

_{3}... x n

_{k}.

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.

_{10}P

_{2}= 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.

###
**User Contributed Comments**
9

You need to log in first to add your comment. ###### smiley25

Multiplication >=2 groups

Factorial, combination, permutation within 1 group.

Factorial, permutation: order does matter

Combination: order does NOT matter

###### atikusz

the above is wrong in the sense that:

combination, permutation considers/applies to 2 groups ! yes, factorial considers 1 group , and labeling ( n!/n1! x n2!..nn!) more than 2 groups.

###### dravinskis

6 lotto numbers to choose out of 49 numbers using the combination formula = 13,983,816 possible combinations. If the jackpot was $56 million you could cover all possible combinations and come out with a lot of money! If only someone would give me a $14M loan.

###### AUAU

dravinskis: If the jackpot was shared by several people then you will lose.

###### Nganeziyamfisa

well explained!!!

###### SergoUA

dravinskis: thank you. good explanation

###### TiredHand

@ dravinskis unfortunately that doesn't include the liklihood of somebody else having the same numbers as you and you having to share the jackpot...

###### hit81

combination: selection

Permutation: arrangement

###### Bududeen

davinskis u are forgeting the power balls combination... if there are 5 to chose 2 from then we have 10 combination... giving a total combination of 140M.. meaning $140M assuming the ticket cost $1 ..thus u lose big time... also need to pay tax in your winnings and also the possibility of other winners at the same time... u are in for a rude awakening!!!!!!