Download 12-1 and 12-2

12-1 and 12-2
Permutations and Combinations
Get out that calculator!! 
If we were to have a class trip to
Great Adventure
and there were only 3 rides (Ferris Wheel, Roller
Coaster and Water Ride), in how many different
orders could you ride them?
6 paths, right? You have 3 to choose from then 2,
then one.
(3 rides, picking one at a time)
Fundamental Principle of Counting
If there are m1 ways to do a first task, m2
ways to do a second task, m3 ways to do a
third task…… mn ways to do the nth task,
then the total possible “patterns” or ways
you could do all the tasks is
m1·m2·m3...mn
Examples
1.
How many 3 digit numbers can be formed if
repetitions are allowed?
2. How many 3 digit numbers can be formed if
repetitions are NOT allowed?
3. How many 4 letter “words” can be made if
duplication of letters is allowed?
4. How many 4 letter “words” can be made if
duplication of letters is not allowed?
Examples
5. How many 3 letter words can be formed in the
first letter must be q? (no repetitions)
6. How many 3 letter words can be formed with
middle letter m if repetitions is allowed?
7. In a game of poker, how many possible 5 card
hands are there?
A new notation
! is called the factorial symbol
n! = n(n-1)(n-2)(n-3)…..1
3! = 3(2)(1) = 6
5! = 5(4)(3)(2)(1)= 120
Permutations
A permutation of “n” elements taken “r” at a
time is an ordered arrangement (without
repetition) of r of the n elements and it is
called nPr.
n!
n Pr 
(n  r)!
The thing to remember is that ORDER
MATTERS!!
Examples
8. How many ways can 4 coins (dime, nickel,
quarter and penny) be arranged in a row?
9. There is a club of 5 boys and 8 girls. The
president will be a girl, the VP will be a boy,
and the secretary and treasurer can each be
either a boy or a girl. How many “councils” are
possible??
What if order doesn’t matter?
What if everyone in a class of 16 had to
shake hands. Using the fundamental
theorem of counting, don’t you get twice
the number of handshakes? Because we
would divide 16P2 by 2 since order doesn’t
matter.
When order doesn’t matter, it is called a
Combination
Consider this question
You and 3 of your friends go to Martinsville
Pizza for an afternoon snack. You decide
to get a large pie. Martinsville pizza offers
5 possible toppings (pepperoni,
mushroom, sausage, green peppers and
onion). How many possible pizza orders
could be made?
Keep in mind:
You may have a pizza from 1 to 5 toppings.
Some shorthand
P = pepperoni
M = mushroom
S = sausage
G = green peppers
O = onions
No topping
1 no topping pizza
One topping
P or M or S or G or O = 5 one topping pizzas
Two topping
(Note: Sausage and Mushroom is the same as Mushroom
and Sausage, right?)
P with M, S, G or O (4)
M with S, GP, or O (3)
S with GP or O
(2)
GP with O
(1)
10 two topping pizzas
Three topping
P with M,S or M,G or M,O
with S,G or S,O
with G,O
(6)
M with S,G or S,O
with G,O
(3)
S with G,O
(1)
10 three topping pizzas
Four topping
P with M,S,G or M,S,O or M,G,O or S,G,O
(4)
M with S,G,O
(1)
5 four topping pizzas
Five topping
There’s only one:
P, M, S, G and O.
1 Five topping pizzas
So how many possible pizzas is
this?
1+5+10+10+5+1 = 32 possible pizzas.
Is there an easier way to determine these
numbers without actually writing out the
possibilities? This method seems very
tedious – what if you are at an ice cream
store and have 12 ice cream flavors and 9
toppings?
Lets just see if we can determine a
formula
Look at the 2 topping pizzas.
At first, how many toppings can you choose
from?
How about the next number of choices?
Look at this to see if it makes sense
P
M
M S G O P S G O
S
P M GO
G
O
P M S O P M S G
This looks like 20, which is 5 times 4. Why did
we only have 10?
Because we cancelled the overlap. There were
twice as many pizzas because there were 2 sets
of everything.
Pepperoni/Onion is the same as Onion/Pepperoni.
Since there were two elements, for each
combination, there was the matching one
(in reverse order). So by dividing by the
overlap, we got the true number of
possibilities.
What if there were 10 possible toppings and
we were looking at two topping pizzas?
10•9 = 90 then divide by 2, so 45 possible
pizzas.
What about other numbers of
toppings?
What if you did 3 toppings as if order mattered?
What was the actual number of 3 topping pizzas?
What if you did 4 toppings as if order mattered?
What was the actual number of 4 topping pizzas?
So what must we divide 5P3 by?
What must we divide 5P4 by?
A Combination
A Combination n elements, r at a time, is
given by the symbol
n
n Cr   
r
And can be expanded as
n!
r!(n  r)!
What does the formula do?
n! determines the total number of
(n  r)! possibilities (including overlap)
and dividing by r! takes care of the overlaps.
2! = 2 (with 2 toppings)
3! = 6 (with 3 toppings)
Examples
10. I have a half dollar, penny, nickel, dime and
quarter. How many 3 coin combinations are
there?
11. I have a group of 10 boys, 15 girls. How many
committees of 5 can I create of 2 boys and 3
girls?
12. How many teams of 6 hockey players be
chosen from a group of 12 if position (i.e.
order) doesn’t matter?
Homework
p. 745 11-25 odd
q. 752 11-17 odd