THE FUNDAMENTAL COUNTING PRINCIPLE & PERMUTATIONS

THE FUNDAMENTAL COUNTING PRINCIPLE &

PERMUTATIONS ESSENTIAL QUESTION

How is the counting principleapplied 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 meatsEvent 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 meats3 cheeses3 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.

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

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 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

nPr = !!

rn

n

BACK TO THE LAST PROBLEM WITH THE SKIERSIt 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 visit6 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