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9.6 The Fundamental Counting Principal & Permutations

9.6 The Fundamental Counting Principal & Permutations

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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

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Page 1: 9.6 The Fundamental Counting Principal & Permutations

9.6The Fundamental Counting Principal & Permutations

Page 2: 9.6 The Fundamental Counting Principal & Permutations

The Fundamental Counting Principal

• 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

Page 3: 9.6 The Fundamental Counting Principal & Permutations

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

Page 4: 9.6 The Fundamental Counting Principal & Permutations

• At a restaurant at Cedar Point, you have the choice of 8 different entrees, 2 different salads, 12 different drinks, & 6 different desserts.

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

• 8*2*12*6=• 1152 different dinners

Page 5: 9.6 The Fundamental Counting Principal & Permutations

Fund. Counting Principle with repetition

• Ohio License 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

Page 6: 9.6 The Fundamental Counting Principal & Permutations

How many plates are possible if digits and letters 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

Page 7: 9.6 The Fundamental Counting Principal & Permutations

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

Page 8: 9.6 The Fundamental Counting Principal & Permutations

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

Page 9: 9.6 The Fundamental Counting Principal & Permutations

Using Permutations

• An ordering of n objects is a permutation of the objects.

Page 10: 9.6 The Fundamental Counting Principal & 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

Page 11: 9.6 The Fundamental Counting Principal & Permutations

In general, the # of permutations of n objects is:

•n! = n*(n-1)*(n-2)* …

Page 12: 9.6 The Fundamental Counting Principal & Permutations

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

Page 13: 9.6 The Fundamental Counting Principal & Permutations

Factorial with a calculator:

•Hit math then over, over, over.•Option 4

Page 14: 9.6 The Fundamental Counting Principal & Permutations

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

Page 15: 9.6 The Fundamental Counting Principal & Permutations

Permutation of n objects taken r at a time

•nPr = !!rnn

Page 16: 9.6 The Fundamental Counting Principal & Permutations

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

Page 17: 9.6 The Fundamental Counting Principal & Permutations

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

Page 18: 9.6 The Fundamental Counting Principal & Permutations

How many ways can you visitall 10 of them:

• 10P10 = • 10!/(10-10)! = • 10!/0!=• 10! = ( 0! By definition = 1)• 3,628,800

Page 19: 9.6 The Fundamental Counting Principal & Permutations

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

Page 20: 9.6 The Fundamental Counting Principal & Permutations

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!

Page 21: 9.6 The Fundamental Counting Principal & Permutations

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

Page 22: 9.6 The Fundamental Counting Principal & Permutations

Find the number of distinguishable permutations of the letters:

• SUMMER :

• 360

• WATERFALL :

• 90,720

Page 23: 9.6 The Fundamental Counting Principal & Permutations

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

possible

• 8!/(3!*5!) = • 56