Transcript File

pencil, highlighter, GP notebook,
calculator, red pen
You go to a store called “40 Flavors.” In
how many ways can you order:
a) a two scoop cone
(with repeats)
b) two scoop cone
(without repeats)
40 P2 
______
40 40 = 1600
+2
+1
c) 3 scoop cone
(with repeats)
_________
40 40 40 = 64,000
+2
+1
+2
40!
40!

 40  39
40  2 ! 38!
 1560 +1
d) 3 scoop cone
(without repeats)
+2
40!
40!
40 P3 

40  3  ! 37!  40  39  38
 59,280
+1
total:
12
Recall:
order
repetition
Permutations – __________
matters and no ____________.
nPr
r things from ___
n things.
is choosing and arranging ___
Example #1: You go to a store called “27 Flavors.” In how
many ways can you order:
a) a two scoop cone
(with repeats)
______
27 27 =
729
b) two scoop cone
(without repeats)
27 P2

27!
27!
 27  26

27  2 ! 25!
 702
c) 3 scoop cone
(with repeats)
_________
27 27 27 = 19,683
d) 3 scoop cone
(without repeats)
27!
27!
27 P3 

27  3  ! 24!
 27  26  25
 17,550
e) A two scoop bowl (without repeats)
Does order matter?
NO
Are repeats allowed?
NO
YES
Combination Leave blank
YES
Are repeats allowed?
NO
Permutation
YES
Decision Chart
According to our flowchart, this is a combination.
So how do we figure this one out?
order in a bowl. This is
Part e: There is no ________
combination In how many ways can you order a
called a _____________.
two scoop bowl (without repeats)?
This is NOT a permutation. If it was, how would you
calculate the possibilities?
27!
27!
27 P2  27  2  !  25!  27  26  702
But since order does NOT matter, we have repeats. This is
how to get rid of the repeats:
2!
 351
Will it always be 2! ? Where did the two come from?
Let us look at an example with less possibilities.
Example #2: Amy, Betty, Carla, Danielle, Erin, and Francis all
try out for 2 places on the cheer squad.
a) Suppose there were two positions (cheer leader and co–leader)
that the girls were trying out for. How many different ways can
the positions be filled? This is a permutation!
6!
6!
 30

6

5

P

6 2
6  2 ! 4!
b) Suppose that the girls were going to be on a cheer committee.
Write out all the possible ways to choose 2 different girls.
(Be methodical!)
This is a combination!
AB
AC
AD
AE
AF
BC CD DE EF
BD CE DF
BE CF
BF
What about BA, CA . . . ?
These don’t matter since
order doesn’t matter.
15
# of combinations: ______
permutation In part
Since order does not matter, this is not a ____________.
(b), we care about who is selected but we do not care about the
order of the selection or any arrangement of the groups.
Selections of committees, or of subsets of items from a larger set
without regard to the order of the group selected, are called
combinations
______________.
In example 2 part b,
15
# of permutations = ____
30 = ____
6 P2 = _____
# of combinations = _________________
2!
2
# of arrangements
IC – 51
To win the California lottery, a person must choose the correct
combination of numbers; that is, choosing the correct 6 of 51
____________
order
numbers. (_________
does not matter.)
a) How many combinations of 6 numbers can be chosen from
the list of 51?
Total # of ways to arrange 6 #s
out of 51 #s when order matters.
51P6 = _______________ = 18,009,460
# of combinations = _______
_________
6!
720
How many ways are there
to arrange 6 numbers?
51!
51!


 51  50  49  48  47  46  12,966,811 ,200
(51  6)! 45!
b) What is the probability of winning the lottery by choosing just
1
1 set of 6 numbers?
P(winning) 
18,009,460
Complete IC – 52 in your groups.
IC – 52
Five cards are drawn from a standard deck of 52 cards.
a) Use permutation notation to calculate the number of choices
for the 1st card, 2nd card, and so forth up to the fifth card.
b) How many ways can you arrange the five cards selected?
Write your answer both as a number and using a factorial.
c) Since order generally does not matter when playing cards, we
need to divide out the number of repetitions of the same set of
five cards. Calculate the number of five card hands that can be
selected from a deck of 52 cards.
# of combinations =
IC – 52
Five cards are drawn from a standard deck of 52 cards.
a) Use permutation notation to calculate the number of choices
for the 1st card, 2nd card, and so forth up to the fifth card.
52!
52!

 52  51  50  49  48  311,875,20 0
52 P5 
(52  5)! 47!
b) How many ways can you arrange the five cards selected?
Write your answer both as a number and using a factorial.
5!  5  4  3  2  1  120
c) Since order generally does not matter when playing cards, we
need to divide out the number of repetitions of the same set of
five cards. Calculate the number of five card hands that can be
selected from a deck of 52 cards.
52 P5 = ____________
311,875,20 0 = ___________
# of combinations =_______
2,598,960
120
5!
IC – 53
Let’s summarize what we have learned thus far…
arrange a
Factorials  These count the number of ways to ________
order
group of objects in ____________.
choose and _________
arrange
Permutations  These count ways to ________
order matters! These are
a subgroup from a larger group. ______
based on decision charts: how many ways to pick the first object,
times how many ways to pick the second, etc.
nPr
choosing and ___________
arranging r things from n things.
is __________
n pr 
n!
(n – r)!
choose a
Combinations  These count ways to just ________
subgroup from a larger group.
# of ways to choose and arrange
# of combinations = _____________________________
# of ways to arrange
COMBINATIONS
choosing r things from n things. This is when
represents __________
order does NOT matter.
_______
nCr
n!
n!
1
n Pr  (n  r)! 

n Cr 
r!
(n  r)! r!
r!
n!

(n  r)! r!
n!
n Cr 
(n  r)! r!
r! times as many permutations as combinations.
Note: There are __
Remember, you divide by r! because you are getting rid of the
repeats from the permutation. They are now repeats because
with combinations, order does NOT matter.
Add to the flow chart:
Does order matter?
NO
Are repeats allowed?
NO
Combination
nCr
YES
YES
Are repeats allowed?
NO
Permutation
nPr
YES
Decision Chart
nr
Examples:
a) Ten people are running for the prom committee. In how many
different ways can a committee of 4 be chosen?
n!
n Cr 
(n  r)! r!
10!
10!

10 C 4 
(10  4)!4! 6!4!
= 210
b) You go to an ice cream store called “29 Flavors.” How many
different ways can you order a 5 scoop bowl (without repeats)?
29!
29!

= 118,755
29 C 5 
(29  5)!5! 24!5!
Examples:
c) Thirty employees are running for a union committee. In how
many different ways can a committee of 20 be chosen?
n!
n Cr 
(n  r)! r!
30 C 20
30!
30!


= 30,045,015
(30  20)!20! 10!20!
d) Cody has 15 hard boiled eggs. How many different ways can
he select 6 eggs to decorate for a project?
15!
15!

15 C 6 
(15  6)!6! 9!6!
= 5005
Finish the assignment:
IC 54 – 60, 63, 64, 66
old bellwork
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calculator, red pen
1. How many different 4 letter passwords can you have if
you can not have repeats and you can only use the letters
g through p?
total:
4
How many different 4 letter passwords can
you have if you can not have repeats and
you can only use the letters g through p?
g, h, I, j, k, l, m, n, o, p
10!
10!


 10  9  8  7  5040
10 P4
(10  4)!
6!
+2
+2
There are 5040 passwords.
total:
4