The Fundamental Counting Principle & Permutations
Computer Science, Statistics and Probability all involve counting techniques which are a branch of mathematics called combinatorics (ways to combine things). We'll be introducing this topic in this section.
For dinner you have the following choices:
soup salad chicken hamburgerprawns
icecream
ENTREES MAINS
DESSERTSHow many different combinations of meals could you make?
We'll build a tree diagram to show all of the choices.
soup
salad
chicken
prawnshamburger
chicken
hamburger
prawns
ice cream
ice cream
ice cream
ice cream
ice cream
ice cream
Now to get all possible choices we follow each path.
soup, chicken, ice cream
soup, chicken,
soup, prawns, ice cream
soup, prawns,
soup, hamburger, ice cream
soup, hamburger,
salad, chicken, ice creamsalad, chicken,
salad, prawns, ice cream
salad, prawns,
salad, hamburger, ice cream
salad, hamburger,
Notice the number of choices at each branch
2choices
3choices
2choices
We ended up with 12 possibilities
2 3 2 = 12
The Fundamental Counting Principle & Permutations
Essential Question
How is the counting principleapplied to determine
outcomes?
Multiplication Principle of CountingIf a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in
different ways.
p q r
If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear?
6 4 5 3 = 360
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
Top Related