10.1- The Fundamental Counting Principal- Permutations- Factorials

The Fundamental Counting PrincipleThe 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 breadsHow 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.

Question: How many different dinners can you choose (one choice of each)?

8*2*12*6=

1152 different dinners

Fundamental Counting Principle with Repetition

Let’s say, Ohio Licenses plates have 3 #’s followed by 3 letters. 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?

Same situation with license plates…but a number or letter 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

PASSWORDS

Let’s pick a password….

a) 4 using digits only

PASSWORDS

Let’s pick a password….

b) 4 using digits or letters (not case sensitive)

PASSWORDS

Let’s pick a password….

c) 4 using digits or letters (upper/lower case sensitive)

TestingA 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 Fund. 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

•nPr = !

!

rn

n

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 visitall 10 of them:

• 10P10 =

• 10!/(10-10)! =

• 10!/0!=

• 10! = ( 0! By definition = 1)

• 3,628,800

How many different ways can we arrange the letters A, B, &C

• ABC

• ACB

• BAC

• BCA

• CAB

• CBA

3!

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,MMOM, OMM, MMOMOM, OMM, MMOOnly 3 are Distinguishable.

!2

!3 3

Find the number of Find the number of Distinguishable Distinguishable permutations of the letters:permutations of the letters:

• OHIO : 4 letters with O 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

SUMMER

360

WATERFALL

90,720

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!

A dog has 8 puppies, 3 male and 5 female. How many birth orders are

possible by gender.

• 8!/(3!*5!) =

• 56

Assignment