Transcript 12.1 The Fundamental Counting Principal

The Fundamental Counting
Principle & Permutations
Why do you use a fundamental counting principal?
What operation do you use for fundamental counting
principals?
What is a permutation?
What is the formula for nPr?
What is the formula for permutations with repetition?
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 different 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
Fund. 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.
Introduction to Permutations
http://www.khanacademy.org/math/probabilit
y/v/permutations
There are 6 permutations of the
letters A, B, &C
•
•
•
•
•
•
ABC
ACB
BAC
BCA
CAB
CBA
You can use the
Fund. Counting Principal 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
!
•nPr =
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
So far in our problems, we have
used distinct objects.
• If some of the objects are repeated, then
some of the permutations are not
distinguishable.
• There are 6 ways to order the letters
M,O,M
• MOM, OMM, MMO
• MOM, OMM, MMO
• Only 3 are distinguishable. 3!/2! = 6/2 = 3
Permutations with Repetition
• The number of DISTINGUISHABLE
permutations of n objects where one
object is repeated q1 times, another is
repeated q2 times, and so on :
n!
q1! q2 ! ...  qk !
Find the number of distinguishable
permutations of the letters:
• OHIO : 4 letters with 0 repeated 2 times
• 4! = 24 = 12
• 2! 2
• MISSISSIPPI : 11 letters with I repeated 4
times, S repeated 4 times, P repeated 2
times
•
11!
= 39,916,800 = 34,650
• 4!*4!*2!
24*24*2
Find the number of distinguishable
permutations of the letters:
• SUMMER :
• 360
• WATERFALL :
• 90,720
A dog has 8 puppies, 3 male and 5
female. How many birth orders are
possible
• 8!/(3!*5!) =
• 56
Why do you use a fundamental counting principal?
To count the number of possibilities of the given conditions.
What operation do you use for fundamental counting
principals?
Multiplication
What is a permutation?
An ordering of objects.
n!
What is the formula for nPr? n  r !
What is the formula for permutations with repetition?
n!
q1! q2 ! ...  qk !
Assignment
worksheet