Transcript Counting Principle and Permutations PowerPoint

THE FUNDAMENTAL
COUNTING PRINCIPLE
& PERMUTATIONS
THE FUNDAMENTAL COUNTING
PRINCIPLE & PERMUTATIONS
ESSENTIAL QUESTION
How is the counting principle
applied to determine outcomes?
THE FUNDAMENTAL COUNTING
PRINCIPLE
If you have 2 events: 1 event can occur m ways
and another event can occur n ways, then the
number of ways that both can occur is m*n
Event 1 = 4 types of meats
Event 2 = 3 types of bread
How many diff types of sandwiches can you
make?
4*3 = 12
3 OR MORE EVENTS:
3 events can occur m, n, & p ways, then the
number of ways all three can occur is
m*n*p
4 meats
3 cheeses
3 breads
How many different sandwiches can you
make?
4*3*3 = 36 sandwiches
At a restaurant at Cedar Point, you have
the choice of 8 different entrees, 2
different salads, 12 different drinks, & 6
different deserts.
How many different dinners (one choice of
each) can you choose?
8*2*12*6=
1152 different dinners
FUNDAMENTAL COUNTING
PRINCIPLE WITH REPETITION
Ohio Licenses plates have 3 #’s followed
by 3 letters.
1. How many different licenses plates are
possible if digits and letters can be
repeated?
There are 10 choices for digits and 26
choices for letters.
10*10*10*26*26*26=
17,576,000 different plates
HOW MANY PLATES ARE POSSIBLE
IF DIGITS AND NUMBERS CANNOT
BE REPEATED?
There are still 10 choices for the 1st digit
but only 9 choices for the 2nd, and 8 for
the 3rd.
For the letters, there are 26 for the first,
but only 25 for the 2nd and 24 for the 3rd.
10*9*8*26*25*24=
11,232,000 plates
PHONE NUMBERS
How many different 7 digit phone
numbers are possible if the 1st
digit cannot be a 0 or 1?
8*10*10*10*10*10*10=
8,000,000 different numbers
TESTING
A multiple choice test has 10
questions with 4 answers each.
How many ways can you
complete the test?
4*4*4*4*4*4*4*4*4*4 = 410 =
1,048,576
USING PERMUTATIONS
An ordering of n objects
is a permutation of the
objects.
THERE ARE 6 PERMUTATIONS OF
THE LETTERS A, B, &C
ABC
 ACB
 BAC
 BCA
 CAB
 CBA

You can use the Fundamental
Counting Principle to determine
the number of permutations of n
objects.
Like this ABC.
There are 3 choices for 1st #
2 choices for 2nd #
1 choice for 3rd.
3*2*1 = 6 ways to arrange the
letters
IN GENERAL, THE # OF
PERMUTATIONS OF N OBJECTS
IS:
n! = n*(n-1)*(n-2)*
…
12 SKIERS…
How many different ways can 12 skiers in the
Olympic finals finish the competition? (if there
are no ties)
12! =
12*11*10*9*8*7*6*5*4*3*2*1 =
479,001,600 different ways
FACTORIAL WITH A
CALCULATOR:
•Hit math then over, over,
over.
•Option 4
BACK TO THE FINALS IN THE
OLYMPIC SKIING COMPETITION
How many different ways can 3 of the
skiers finish 1st, 2nd, & 3rd (gold,
silver, bronze)
Any of the 12 skiers can finish 1st, the
any of the remaining 11 can finish
2nd, and any of the remaining 10 can
finish 3rd.
So the number of ways the skiers can
win the medals is
12*11*10 = 1320
PERMUTATION OF N OBJECTS
TAKEN R AT A TIME
n!
P
=
n r
n  r !
BACK TO THE LAST PROBLEM
WITH THE SKIERS
It can be set up as the number of permutations of 12
objects taken 3 at a time.
12P3
= 12!
= 12! =
(12-3)!
9!
12*11*10*9*8*7*6*5*4*3*2*1 =
9*8*7*6*5*4*3*2*1
12*11*10 = 1320
10 COLLEGES, YOU WANT TO
VISIT ALL OR SOME
How many ways can you visit
6 of them:
Permutation of 10 objects taken 6
at a time:
10P6 = 10!/(10-6)! = 10!/4! =
3,628,800/24 = 151,200
HOW MANY WAYS CAN YOU
VISIT ALL 10 OF THEM:
10P10
=
10!/(10-10)! =
10!/0!=
10! = ( 0! By definition = 1)
3,628,800