Transcript Fundamental Counting Principle and Permutations

Formulas and Principles
Math I Unit 4
If one event can occur in m ways and another
event can occur in n ways, then the number
of ways that both events can occur is …
m n
 You
are buying a sandwich. You have a
choice of 5 meats, 4 cheeses, 3 dressings,
and 8 other toppings. How many
different sandwiches with one meat, one
cheese, one dressing, and one other
topping can you choose?
 You
can use the fundamental counting
principle to find the total number of
sandwiches.
Number of sandwiches = 5 x 4 x 3 x 8
= 480
A
town has telephone numbers that
begin with 432 or 437 followed by four
digits. How many different telephone
numbers are possible?
 The
first three digits can only be 432 or
437…so there are only 2 choices.
 The last four digits have numbers 0 thru
9 to choose from….so there are 10 choices
for each digit.
432 - ___ ___ ___ ___
437 - ___ ___ ___ ___
2 x 10 x 10 x 10 x 10 =
20,000
Phone number combinations

If you lived in the same town that had 432 and 437 as
the first digits, but the last four digits could not
repeat, how many different telephone numbers can
be created?


2 choices for 432 or 437
10 choices for the first digit
9 choices for the second digit
8 choices for the third digit
7 choices for the fourth digit

Solution: 2 x 10 x 9 x 8 x 7 = 10,080



 Twenty-six
golfers are competing in the
final round of a local competition. How
many different ways can 3 of the golfers
finish first, second, and third?
26 golfers can finish first,
Any of the ____
25 remaining golfers
then any of the ____
can finish second, and finally any of the
24 remaining golfers can finish third.
____
 So, the number of ways that the golfers
can place first, second, and third is…
26 x 25 x 24 = 15,600

 n!
= n x (n - 1) x (n – 2) x (n – 3) x … x 1
Example: Find 5!
5 x 4 x 3 x 2 x 1 = 120
*** 0! Is always equal 1 ***

When finding the number of possible
outcomes where the ORDER MATTERS,
use a permutation.
n is the total number
 r is the number taken at one time

n!
n Pr 
( n  r )!
 You
were left a list of 8 chores. In how
many orders can you complete 6 of the
chores?
 The
number of permutations of 8
chores taken 6 at a time is:
8!
8!
 20,160

8 P6 
(8  6)! 2!
 How
many different orders can you
do all 8 chores?
8
P8
8!
8!
8
!

 
(8  8)! 0! 1
 8!
 40,320
 Twenty-six
golfers are competing in the
final round of a local competition. How
many different ways can 3 of the golfers
finish first, second, and third?

There are 26 golfers to choose from. Since
three will be chosen … and the award that they
receive MATTERS… it can be solved using
permutations.
 26
P 3 = 15, 600

When finding the number of possible outcomes
where the order DOES NOT matter, use a
combination.

n is the total number
r is the number taken at one time


Eighteen basketball players are competing for
5 starting positions. The players selected to
start will make up the first team. If the order
in which the players are selected is not
important, how many different first teams are
possible?
 The
number of ways to choose 5
players from 18 is:
18
!
18
!

18 C5 
((18  5)!5!) (13!5!)
 8568
 TI–84
Plus:
18 MATH PRB (3) nCr 5 ENTER 8568
 TI-30XIIS:
18 PRB nCr ENTER 5
8568

Your English teacher has asked you to select 3
novels from list of 10 to read as an
independent project. In how many ways can
you choose which books to read?
 Since
you have 10
novels to choose
from and you are
choosing only 3…
10
C3  120

Addition Principle:
“OR”
If two or more events cannot occur at the same
time…ADD the combinations together

Multiplication Principle: “AND”
If two or more events can occur at the same
time…MULTIPLY the combinations together
A restaurant serves omelets that can be ordered with
any of the ingredients shown:


Vegetarian: green pepper, red pepper, onion,
mushroom, tomato, cheese
Meat: ham, bacon, sausage, steak
A. Suppose you want exactly 2 vegetarian ingredients
and 1 meat ingredient in your omelet. How many
different types of omelets can you order?


You can choose 2 of 6
vegetarian ingredients
and 1 of 4 meat
ingredients. So, the
number of possible
omelets is:
Since you are choosing
both for the same omelet,
you will multiply the
combinations together.
6
C 2  15
4
C1  4
15  4  60

B. Suppose you can afford at most 3
ingredients in your omelet. How many
different types of omelets can you order?


You can order an omelet
with 0, 1, 2, or 3
ingredients. Because there
are 10 items to choose
from…
You will either be choosing
0 OR 1 OR 2 OR 3 - you
will add the combinations.
10
C0 ,10 C1 ,10 C2 ,10 C3
10
C0 10 C1 10 C2 10 C3
 1 10  45 120  176

A movie rental business is having a special on
new releases. The new releases consist of 8
comedies, 3 family, 10 action, 7 dramas, and 2
mystery movies.
Part 1:
Suppose you want exactly 3 comedies and 2
dramas. How many different movie
combinations can you rent?

You can choose 3 of the 8 comedies and 2 of the
7 dramas. So, the number of possible movie
combinations is:
8
C3
8!
8!


 56
(8  3)!3! 5!3!
7!
7 C2 
(7  2)!2!
7!

5!2!
 21
56 x 21 = 1176

Suppose you can afford at most 2 movies. How
many movie combinations can you rent
(assuming that you do rent a tleast 1 movie)?

Since you can either
get 1 movie OR 2
movies…
C1
 30
C2
 435
30
30
30  435  465