1. How to Count Things
“There are three kinds of people in the world: those who can count and those who cannot.”
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2. Why is counting important?
Recall: the basic definition of the probability of an event, 𝑃 𝐸 = 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑡ℎ𝑎𝑡 𝑓𝑖𝑡 𝐸𝑣𝑒𝑛𝑡 𝐸 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
You need to be able to count how many outcomes fit the event of interest.
You need to be able to count how many outcomes are in the sample space.
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3. Counting can get complicated
There are techniques and tricks to counting
For instance: probability of drawing five cards and getting a Full House
How many different poker hands can be drawn?
How many of them are Full House hands?
Making a list of the sample space outcomes would take a very, very long time!
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4. Where we’ve been; Where we’re going
Previously:
Elementary probability
P(A or B), the addition rule
P(A and B), the multiplication rules
Now:
Counting events, just counting, only
Using counting in P(E) computations that are more challenging than the ones we’ve been doing
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5. Three Counting Tricks
The Fundamental Counting Rule – how many different license plates can be made?
Permutations – ten horses race; how many different win/place/show results are possible?
Combinations – how many different juries can be formed from the 100 jurors in today’s pool?
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6. The Fundamental Counting Principle
The setup:
A sequence of 𝑛 events
There are 𝑘 1 possible outcomes of event #1
There are 𝑘 2 possible outcomes for event #2
Etc., etc., 𝑘 𝑛 possible outcomes for event #𝑛
The conclusion: You Multiply their counts.
Overall 𝑘 1 ∙ 𝑘 2 ∙⋯∙ 𝑘 𝑛 possible outcomes
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7. Classic Example - License Plate
Suppose we must have three letters followed by four numbers.
How many possible different license plates?
Use the Fundamental Counting Principle!
letter letter letter number number number number
____ ∙ ____∙ ____∙ ____ ∙ ____ ∙ ____ ∙ ____
= ___________ possible different license plates
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8. Classic Example – The Menu
You get one appetizer out of six choices,
And one main course out of ten choices,
And one dessert out of eight choices.
How many meals before you’ve tried every possible combination?
Appetizer and Main Course and Dessert
________ ∙ _________ ∙ _______
= ___________
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9. Classic Example – The Meeting Room
A, B, C, D, E, F, X, Y, Z attend a meeting
The chairman will sit at the head of the table.
Only X, Y, and Z are qualified to be chairman.
There are four seats along each side of the table.
How many possible seating arrangements are there? (Hint: Start with chairman, then fill the rest of the seats.) _____ ∙ _____ ∙ _____ ∙ _____ ∙ _____ ∙ _____ ∙ _____ ∙ _____ ∙ _____
= ______________ possible different seatings
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10. Factorials Example: 10! It means 10∙9∙8∙7∙6∙5∙4∙3∙2∙1
The result is a large number!
TI-84 MATH, PRB, 4:!
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11. Permutations The situation The notation and the formula
You have 𝑛 objects
You are going to take 𝑟 of them
The order of the 𝑟 items is important
The notation and the formula
𝑛 𝑃 𝑟 = 𝑛! (𝑛−𝑟)!
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12. Permutations Example 𝑛 𝑃 𝑟 = 𝑛! (𝑛−𝑟)! Ten horses run the race. 𝑛=10
𝑛 𝑃 𝑟 = 𝑛! (𝑛−𝑟)!
Example
Ten horses run the race. 𝑛=10
How many win/place/show possibilities? 𝑟=3
10 𝑃 3 = 10! (10−3)! = 10∙9∙8∙7∙6∙5∙4∙3∙2∙ ∙6∙5∙4∙3∙2∙1
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13. Permutations Example Ten horses run the race. 𝑛=10
How many win/place/show possibilities? 𝑟=3
10 𝑃 3 = 10! (10−3)! = 10∙9∙8∙7∙6∙5∙4∙3∙2∙ ∙6∙5∙4∙3∙2∙1 =10∙9∙8=720
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14. Permutations Example Ten horses run the race. 𝑛=10
How many win/place/show possibilities? 𝑟=3
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15. Permutations
Example
Ten horses run the race. 𝑛=10
How many win/place/show possibilities? 𝑟=3
This happened to be the same as if we used the Fundamental Counting Rule
10 horses could cross the line first and Win
9 remaining horses for the 2nd Place finish
8 remaining horses who could Show
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16. Example: How many “words”
How many four-letter words can be made, if we don’t repeat any letters?
Thinking is the only way to do these !
26 letters
Choose 4 of them
The order matters
So Permutation is the way to do it
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17. How many four-letter “words” ?
____ P _____ = ______
If we instead used the Fundamental Counting Principle to compute this, _____ ∙ _____ ∙ _____ ∙ _____ = _____
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18. How many words?
How many 7-letter “words” can be made from the letters in FLORIDA ?
There are ____ letters available
We are taking ____ of them
Order { does or does not? } matter
So it’s a ___________ situation.
The computation is ___ P ____
The solution is __________.
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19. How many words? How many “words” can be made from FLORIDA ?
This is a multistep problem!
How many 2-letter words? ____ P ____
How many 3-letter words? ____ P ____
How many 4-letter words? ____ P ____
How many 5-letter words? ____ P ____
How many 6-letter words? ____ P ____
How many 7-letter words? ____ P ____
How many 8-letter words? ____ P ____
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20. “Special” Permutations
How many 4-letter words can be made from the letters in the BABY?
Complicated by 2 B’s. Don’t double-count.
Just doing 4! won’t work.
How: “Divide out” the duplicate letters ! 2!
ABBY, ABYB, AYBB,
BABY, BAYB, BBAY, BBYA, BYAB, BYBA,
YABB, YBAB, YBBA
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21. “Special” Permutations
How many 11-letter words can be made from the 11 letters in MISSISSIPPI ?
Complicated by 4 I’s, 4 S’s, 2 P’s
Just doing 11! won’t work.
“Divide out” the duplicate letters. 11! 4!4!2!
EXTRA PARENS NEEDED ON TI-84 !!!
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22. Permutations vs. Combinations
You have 𝑛 objects
You are going to take 𝑟 of them
The order of the 𝑟 items is important
𝑛 𝑃 𝑟 = 𝑛! (𝑛−𝑟)!
You have 𝑛 objects
You are going to take 𝑟 of them
The order of the 𝒓 items is NOT important
So we divide out the various repetitions of those 𝒓 items.
𝑛 𝐶 𝑟 = 𝑛! 𝑛−𝑟 ! 𝒓!
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23. Permutations – Classic Example
How many different 5-card poker hands?
Analysis:
𝑛=52 items
𝑟=5 of them taken at a time
Order does not matter. An ace is an ace, whether you drew it first or second or third or fourth or last
Compute:
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24. Poker Hands Combination
𝑛 𝐶 𝑟 = 𝑛! 𝑛−𝑟 ! 𝑟!
52 𝐶 5 = 52! 52−5 !5! = 𝟓𝟐∙𝟓𝟏∙𝟓𝟎∙𝟒𝟗∙𝟒𝟖∙𝟒𝟕! 𝟒𝟕!∙𝟓∙𝟒∙𝟑∙𝟐∙𝟏
Clever set-up: the 47! top cancels with 47! Bottom
Or TI-84 MATH, PRB, 3:nCr
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25. Classic Example - Committees
Choosing 5 students for the cafeteria committee out of 20 students available.
Analysis
Order doesn’t matter
So it’s a Combination (not a Permutation)
Computation: ___ C ___ =
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26. Classic Multistep Committee Problem
We’re choosing the Scheduling Committee.
10 women and we need 3 of them to serve.
6 men and we need 2 of them to serve.
How many possible committees can we have?
You’ll need two different tools to compute this.
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27. Classic Multistep Committee Problem
Women and Men on the committee
Analysis
Fundamental Counting Principle, multiply:
How many ways to choose 3 out of 10 women?
Times how many ways to choose 2 out of 6 men?
The women are one nCr combination calculation.
The men are another nCr combination calculation.
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28. Classic Multistep Committee Problem
3 Women and 2 Men on the committee
How many ways to choose a President and a Vice-President, with no restrictions on who may serve?
Analysis
Order matters here!
So it’s a Permutation: ___ P ___ = _____
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29. Lottery – How many ways to choose five numbers out of ____ numbers?
5 numbers out of 5? ____ C ____ = ______
5 numbers out of 6? ____ C ____ = ______
5 numbers out of 7? ____ C ____ = ______
5 numbers out of 10? ____ C ____ = ______
5 numbers out of 20? ____ C ____ = ______
5 numbers out of 30? ____ C ____ = ______
5 numbers out of 59? ____ C ____ = ______
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30. Powerball Five white numbers from 1 thru 59
One red Powerball number from 1 thru 39
The Fundamental Counting Principle again
Combination calculation for 5 white numbers
Trivial calculation for the 1 powerball number
Multiply those results together
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31. How many “pairs” in a deck of cards?
View it as two events in a single draw
The rank
The suit
Then apply the Fundamental Counting Princ.
Any one of the 13 ranks (from 2, 3, …, Q, K, A)
Four cards of that rank taken two at a time
13 𝐶 1 ∙ 4 𝐶 2 =78
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32. How many 5-card flushes? View it in two pieces
1 out of the 4 suits
5 out of the 13 cards in that suit
The Fundamental Counting Principle, again
4 𝐶 1 ∙ 13 𝐶 5 =5148
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