Transcript Fundamental counting principle Method two

Warm – Up
Evaluate
1) 3!
March 5, 2010
= 6
2) 7! = 5040
4!
= 1/30
3)
6!
10!
4)
10  4!
= 5040
Quote
It is easier to gain
forgiveness than to
get permission.
Grace Murray Hopper
Puzzle – What is this?
The maker doesn’t want
it, the buyer doesn’t use
it and the user doesn’t
see it.
What is it?
Fundamental counting principle
Fundamental Counting Principle =
Fancy way of describing how we can
determine the number of ways a
sequence of events can take place.
Fundamental counting principle
You are at your school cafeteria that allows you to choose a
lunch meal from a set menu. You have two choices for the
Main course (a hamburger or a pizza), Two choices of a drink
(orange juice, apple juice) and Three choices of dessert (pie,
ice cream, jello).
12 meals
How many different meal combos can you select?_________
Method one: Tree diagram
Lunch
Hamburger
Apple
Pie
Icecream
Jello
Orange
Pie
Icecream
Jello
Pizza
Apple
Pie
Icecream
Jello
Orange
Pie
Icecream
Jello
Fundamental counting principle
Method two: Multiply number of choices
2x2x3=
12 meals
Ex 2: No repetition
During the Olympic 400m sprint, there are 6 runners.
How many possible ways are there to award first,
second, and third places?
3 places
1st
2nd
3rd
____
6 x ____
5 x ____
4 = 120 different ways
Fundamental counting principle
Ex 3: With repetition
License Plates for cars are labeled with 3 letters followed
by 3 digits. (In this case, digits refer to digits 0 - 9. If a
question asks for numbers, its 1 - 9 because 0 isn't really a
number) How many possible plates are there? You can use
the same number more than once.
___
26 x ___
26 x ___
26 x ___
10 x ___
10 x ___
10 = 17,576,000 plates
Ex 4: Account numbers for Century Oil Company consist
of five digits. If the first digit cannot be a 0 or 1, how many
account numbers are possible?
___
8 x ___
10 x ___
10 x ___
10 x ___
10 = 80,000 different account #’s
Factorials
5 • 4 • 3 • 2 • 1 = 5! Factorial
7!= 7 • 6 • 5 • 4 • 3 • 2 •1 = 5040
7! 7  6  5  4  3  2  1

 42
5!
5  4  3  2 1
8! 8  7  6  5  4  3  2  1 8  7  6


5!3! 5  4  3  2  1  3  2  1 3  2  1
87
 56
1
Permutations
Permutations = A listing in which order IS important.
6P4
P(6,4) Represents the number of ways 6 items can be
taken 4 at a time…..
Can be written as:
P(6,4)
Or 6 x 5 x 4 x 3 = 360
or
Or 6 (6-1) (6-2) (6-3)
In general, the number of permutations of n distinct
objects is:
n!  n  n  1  n  2 ..... 3 2 1
Twelve skiers are competing in the final round of Olympic
freestyle skiing aerial competition.
a) In how many different ways can the skiers finish the
competition? (Assume there are no ties.)
There are 12! different ways that the
skiers can finish the competition.
Use your calculators to find 12! = 479,001,600
Permutations of n objects taken r at a time
The number of permutations of r objects taken
from a group of n distinct objects is denoted by
P
n r
and is given by:
n!
n Pr 
 n  r !
You are considering 10 different colleges. Before you
decide to apply to the colleges, you want to visit some or
all of them. In how many orders can you visit (a) 6 of the
colleges and (b) all 10 colleges?
a) The number of permutations of 10 objects taken 6 at a
time is:
10
P6
10!
10! 3, 628,800
 151, 200



24
10  6 ! 4!
b) The number of permutations of 10 objects taken 10 at
a time is:
P
10 10
Use your calculator…
 3, 628,800
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 is:
n!
q1 ! q2 ! ..... qk !
Find the number of distinguishable permutations of the
letters in (a) OHIO and (b) MISSISSIPPI
a) OHIO has 4 letters of which O is repeated 2 times. So,
the number of distinguishable permutations is …
4!
2!
24

2
 12
b) MISSISSIPI has 11 letters of which I is repeated 4
times, S is repeated $ times and P is repeated 2 times. So,
the number of distinguishable permutations is …
11!
39,916,800

4! 4! 2!
24 24 2
 34, 650