COUNTING
By Patrick, Andrew, Jared & Jonathan
COUNTING

Counting is the study of the enumeration of
discrete, finite sets (“counting objects”)


Ex. If there are 5 symbols with which we are given, to
create a 4 symbol license plate, how many plates can we
make? (Symbols can be repeated)
(5)(5)(5)(5) = 625
MULTIPLICATION PRINCIPLE FOR
COUNTING


If an activity can be performed as a sequence of “k”
independent steps, and step “i” can be accomplished in
“ni” ways, than the entire activity can be performed in
(n1)(n2)(n3)…(nk) ways.
When can this be used?

Ex. If we must create a license plate in which two of the
symbols can be of a given set of 6 letters or two of the
symbols can be of a given set of 5 numbers, how many
unique plates can we make?

(5)(5) + (6)(6) = 61
ADDITION PRINCIPLE FOR COUNTING

If an activity can be divided into “k” disjoint
subsets of activities (separate activities), and the
ith of these sets contains ni elements, than total
number is n1+ n2+ n3+….+ nk.
PERMUTATION
A permutation of “n” distinct objects x1, x2, x3…xn
is any ordering of “n” objects.
 P(n,r) = n!/(n-r)! where one chooses “r” items from
“n” items (order matters).
 Where does it come from?

EXAMPLE
If there are 16 baseball players who need to be
put into a 3 player batting order, how many
different batting orders are possible?
 P(16,3) = 16!/(16-3)! = 16!/13! = 3,360.
 There are 3,360 unique batting orders which are
possible.

COMBINATIONS
Combinations are Permutations where the order
of the objects being counted doesn’t matter.
 For example: 1, 2, and 3 can be arranged in 6
different ways if order matters but only 1 way if
order does matter:

Matters
1,2,3
1,3,2
2,1,3
2,3,1
3,2,1
3,1,2
Doesn’t Matter
1,2,3
COMBINATIONS CONT’D



The table from the previous page shows that for a
set of (1,2,3) if you want to order the numbers
with a Permutation you will get 6 outputs, but
with a Combination, you only receive 1.
This shows that there are 6 times as many
Permutations than combinations for this set.
This works for any P(n,r) and C(n,r).
Since there are 3 digits, you divide the number of
permutations by another 3! which is equal to 6.
COMBINATIONS EQUATION


To reach this conclusion we reduce the formula
by how many ways the objects could be in order
which accounts for the r! on the bottom.
P(n,r) = C(n,r)*r!
COMBINATIONS WITH REPETITION

Combinations with repetition are also referred to
as “Multichoose” problems.

The equation is

How does it work?
MULTICHOOSE


Let’s say you have 5 different flavors of ice
cream, and you can have 3 scoops. Your 3 scoops
can be any flavor, with repetition, how many
different combinations of ice cream scoops can
you have?
We will say the flavors are banana, chocolate,
lemon, strawberry and vanilla.
MULTICHOOSE CONT’D


Think about the ice cream being in containers,
you could say "skip the first, then 3 scoops, then
skip the next 3 containers" and you will end up
with 3 scoops of chocolate!
So there are r + (n-1) positions, and we want to
choose r of them to have circles.
EXAMPLE
If we refer back to the previous problem about
batting orders, we will take the same number of
players, 16, but this time 3 players will be picked
to be team captains. How many ways can they be
chosen in groups of 3?
 C(16,3) = 16!/3!(16-3)!
= 16!/3!(13)!
= 16*15*14/3*2*1
= 560
 In this case, there are only 560 ways that players
can be chosen, because order does not matter.

IDENTITY 130
Identity: For 0 ≤ k ≤ n
k(nk)=n(n-1k-1)
 Question: How many ways can we create a size
k committee of students from a class of n
students, where one of the committee members is
designated as the chair?

IDENTITY 130 CONTINUED
There are (nk) ways to choose the committee, then
k ways to select a chair. Therefore, there are k(nk)
possible outcomes.
 First, select the chair from the class of n
students. Then, from the remaining n-1 students,
pick the remaining k-1 members. This can be
done n(n-1k-1) ways.

IDENTITY 131



We have a question we could ask:
How many ways can we make a committee (of any size)
from a class of students of n students, where one of them is
designated as the chair.
Starting this we will ask how many committees we could
make.

This would be the same as (nk).

We must then factor in the people who may be the chair.

This is k.
IDENTITY 131 CONTINUED


Another way to solve the same thing:
We see that there are n students to be the chair of
committees, so we select the chair of each subcommittee
out of n students.

There are now n-1 students remaining.

It follows that there are 2n-1 ways of filling the remaining
committees.

We now have n*2n-1 ways of choosing different ways of
choosing committees.

If we look at the identity we were given, this helps show
that the answer we got is correct.
THANK YOU TO OUR MENTOR

Greg Warrington!
REFERENCES SLIDE!


Proofs That Really Count; Benjamin/Quinn
Introduction to Discrete Mathematics,
Burgmeier/Kost
HOMEWORK PROBLEMS



We have a baseball team made up of 6 players.
A. How many different 3 person batting orders
can we have? (order matters, Permutation)
B. How many different 2 person groups of team
captains can we have? (order doesn’t matter,
Combination)