In a number of different situations, it is not easy to determine the outcomes of an event by counting them individually. Alternatively, counting techniques that involve permutations and combinations are helpful when calculating theoretical probabilities.

This section will examine methods for determining theoretical probabilities of successive or multiple events.

Permutation? or Combination?

The following flow chart will help determine which formula is suitable for any given question. By simply following a series of "yes" or "no" questions, the appropriate formula can be determined.


Ex. 1 - Using Permutations:



The specific outcome of Mike starting in lane 1 and the other two starting in lane 2 and lane
3 can only happen one way, so n(A) = 1. Therefore,


The probability that Mike will start in the first lane next to his other brothers in lane 2 and 3 is
approximately 0.00101.

Ex. 1(a) - Using Permutations:

Exactly Three People form a line at a grocery store. What is the probability that they will line up in descending order of age? (I.e. oldest, middle and youngest)

→Solution using the blank like method:

n(A): # of ways they will line up in descending order of age, thus:



n(S): # of ways all three can line up in any order, thus:



To calculate the probability that the three people will line up in descending order is:



Ex. 2 - Using Combinations:



However, the group may consist of any 3 members from the group of 13. Thus, n(S) is the total
number of possible outcomes.


The probability of selecting a group consisting of runners only is:


Therefore, the probability of selecting a committee with 3 runners is approximately 0.03497.

b) Either the committee is comprised of runners only, or it isn't. Therefore, the probability of the complement A, P(A'), gives the desired result.


So the probability of selecting a committee not comprised of runners is approximately 0.96503.

Ex. 2(b) – Using Combinations:

A group of 12 people is going out town on Saturday night. The group will take three cars with four people in each car. If they distribute themselves among the cars at random, what is the probability that Chantal and Rafael will be in the same car? (*note: f#: friend)

→Solution using the blank line method:

n(A):# of ways Chantal and Rafael will be in the same car, thus:

n(S):# of ways Chantal and Rafael will be in any one of the three cars, thus:






Ex. 3 - Using the Fundamental Counting Principle:


For the third employee, since there are 363 ways this person can have a different birthday from
the other two, the probability that all three employees have different birthdays is:

Now continue the process with the 12 people in the office:

The probability that at least two people in the office have the same birthday is approximately
0.167.

Key Concepts:

• In probability experiments with many possible outcomes, you can apply the fundamental

counting principle and techniques using permutations and combinations.

• Permutations are useful when order is important in the outcomes; combinations are

useful when order is not important.

References:

“Gliffy: BETA.” Gliffy. San Francisco, California, 2005. 18 Dec 2006.

<http://Gliffy.com/>.

Wgman, Diane. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.