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 sandwichesTRANSCRIPT
9.6The 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
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 desserts.
• How many different dinners (one choice of each) can you choose?
• 8*2*12*6=• 1152 different dinners
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
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
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 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
•nPr = !!rnn
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 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 visitall 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