Computer Science, Statistics and Probability all involve
counting techniques which are a branch of mathematics
called combinatorics (ways to combine things).
For dinner you have the following choices:
soup salad chicken shrimp hamburger
How many different combinations
of meals could you make?
We'll build a tree diagram to
show all of the choices.
Notice the number of choices at each branch We ended up with
2 3 2
choices choices choices soup, chicken, ice cream
2 3 2 = 12 soup, chicken,
shrimp soup, shrimp, ice cream
soup, hamburger, ice cream
salad, chicken, ice cream
shrimp salad, shrimp, ice cream
Now to get all salad, hamburger, ice cream
possible choices we
follow each path.
Multiplication Principle of Counting
If a task consists of a sequence of choices in which
there are p selections for the first choice, q selections for
the second choice, r selections for the third choice, and
so on, then the task of making these selections can be
done in pqr
If we have 6 different shirts, 4 different pants, 5 different pairs
of socks and 3 different pairs of shoes, how many different
outfits could we wear?
6 4 5 3 = 360
A permutation is an ordered arrangement
of r objects chosen from n objects.
For combinations order does not matter but for
permutations it does.
There are three types of permutations.
The first is distinct with repetition. This means there are n
distinct objects but in
this means different
choosing r of them you
can repeat an object.
Let's look at a 3 There are 10 choices for the first number
combination lock There are 10 choices for the second number
with numbers 0 and you can repeat the first number
There are 10 choices for the third number
and you can repeat
By the multiplication principle there are 10 10 10 = 1000 choices
This can be generalized as:
Permutations: Distinct Objects with Repetition
The number of ordered arrangements of r objects
chosen from n objects, in which the n objects are
distinct and repetition is allowed, is nr
What if the lock had four choices
for numbers instead of three?
104 = 10 000 choices
The second type of permutation is distinct, without
Let's say four people have a race. Let's look at the
possibilities of how they could place. Once a person has
been listed in a place, you can't use that person again (no
First place would be Based on the multiplication principle:
choosing someone from 4 3 2 1 = 24 choices
among 4 people.
Now there are only 3 to
choose from for second
Now there are only 2
to choose from for
third place. 4th 3rd 2nd 1st
Only one possibility for
nPr, means the number of ordered arrangements of r objects
chosen from n distinct objects and repetition is not allowed.
n r !
4! 4 3 2 1
In the last example:
4 4 ! 0!
If you have 10 people
racing and only 1st, 2nd
and 3rd place how many 0! = 1
possible outcomes are
there? 10! 10 9 8 7!
10 P 720
10 3 ! 7!
A combination is an arrangement of r
objects chosen from n objects regardless
nCr , means the number combinations of r objects chosen
from n distinct objects and repetition is not allowed.
n Cr or
n r ! r! r
Order doesn't matter here so the combination 1, 2, 3 is not
different than 3, 2, 1 because they both contain the same
You need 2 people on your committee and you have 5 to
choose from. You can see that this is without repetition
because you can only choose a person once, and order
doesn’t matter. You need 2 committee members but it
doesn't matter who is chosen first. How many
combinations are there?
5! 5 4 3!
5 C2 10
5 2 !2! 3!2!
The third type of permutation is involving n objects that
are not distinct.
How many different combinations of letters in specific
order (but not necessarily English words) can be formed
using ALL the letters in the word REARRANGE?
E R N R A A G E R
The "words" we form will have 9 letters so we need 9 spots to
place the letters. Notice some of the letters repeat. We need
to use R 3 times, A 2 times, E 2 times and N and G once.
9C3 6C2 4C2 2C1 1C1
84 15 6 2 1 = 15 120 possible "words"
First we choose
That leaves That leaves
positions for the
6 positions 2 positions
R's. There are 9 That leaves
for 2 A's. That leaves for the N.
positions and we'll 1 position
choose 3, order 4 positions
for 2 E's. for the G.
This can be generalized into the following:
Permutations Involving n Objects
That Are Not Distinct
The number of permutations of n objects
of which n1 are of one kind, n2 are of a
second kind, . . ., and nk are of a kth kind
is given by
n1 ! n2 ! nk !
where n n1 n2 nk
A national hamburger chain used to advertise that it
fixed its hamburgers “256 ways” since customers
could order whatever toppings they wanted. How
many toppings must have been available?
For each topping, the choice is yes (put it on my
burger!) or no (leave it off!).
2n = 256
log 2n = log 256
n log 2 = log 256
There must have been 8 possible toppings.
For each of the following numbers, make up
a counting problem that has the number as
• Pg 609 – 611 #11, 13, 17, 19, 25, 35, 39, 45,