Statistical Methods for Psychology

The special branch of mathematics dealing with the number of ways in which objects

can be put together (e.g., the number of different ways of forming a three-person commit-

tee with five people available) is known as combinatorics.Although not many instances in

this book require a knowledge of combinatorics, there are enough of them to make it nec-

essary to briefly define the concepts of permutations and combinations and to give formu-

lae for their calculation.

Permutations

We will start with a simple example that is easily expanded into a more useful and relevant

one. Assume that four people have entered a lottery for ice-cream cones. The names are

placed in a hat and drawn. The person whose name is drawn first wins a double-scoop

cone, the second wins a single-scoop cone, the third wins just the cone, and the fourth wins

nothing. Assume that the people are named Aaron, Barbara, Cathy, and David, abbreviated

A, B, C, and D. The following orders in which the names are drawn are all possible.

ABCD BAC D CA BD D A B C

AB D C B A D C C A D B D A C B

AC B D B C A D C B A D D B A C

AC D B B C D A C B D A D B C A

AD B C B D A C C D A B D C A B

ADCB BDC A CD BA D C B A

Each of these 24 orders presents a unique arrangement (called a permutation) of the

four names taken four at a time. If we represent the number of permutations (arrangements)

of Nthings taken r at a time as , then

where the symbol N! is read Nfactorialand represents the product of all integers from N

to 1. [In other words, By definition, 0! 5 1].

For our example of drawing four names for four entrants,

which agrees with the number of listed permutations.

Now, few people would get very excited about winning a cone without any ice cream

in it, so let’s eliminate that prize. Then out of the four people, only two will win on any

drawing. The order in which those two winners are drawn is still important, however, be-

cause the first person whose name is drawn wins a larger cone. In this case, we have four

names but are drawing only two out of the hat (since the other two are both losers). Thus,

we want to know the number of permutations of four names taken two at a time, ( ). We

can easily write down these permutations and count them:

AB BA CA DA

AC BC CB DB

AD BD CD DC

Or we can calculate the number of permutations directly:

P^42 =

4!

(4–2)!

=

4 # 3 # 2 # 1

2

=12.

P^42

P^44 =

4!

(4 2 4)!

=

4!

0!

=

4 # 3 # 2 # 1

1

= 24

N!=N(N 2 1)(N 2 2)(N 2 3)Á(1).

PNr =

N!

(N 2 r)!

PNr

Section 5.6 Permutations and Combinations 121

combinatorics

permutation

factorial