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Algebra III
Lesson 38
Fundamental Counting Principle and
Permutations – Designated Roots –
Overall Average Rate
Fundamental Counting Principle and Permutations
Using the letters A, B, and C, how many unique combinations
can be made?
List them.
A B C
B A C
A C B
B C A
C A B
C B A
6 different ways to arrange these three letters.
These 6 combinations represent a permutation of the set
of letters of A, B, and C.
Permutation – the arrangement of the members of a set in a
definite order without repetition of set members.
3 2 1
There where three choices of what to put into the first box.
If repetition is not allowed, then there are two choices for the next spot.
And finally, only once choice for the last spot.
Notice: 3 choices times 2 choices times 1 choice = 6 ways
This is an example of the Fundamental Counting Principle
Fundamental Counting Principle – If once choice can be made A
ways and, after the first choice is made, another choice can be made
in B ways, then the number of possible choices, in order, is A times B
different ways.
What if repetition were allowed of A, B, or C, how many choices
would there be?
3 3 3
How many options for the first box?
How many options for the second box?
How many options for the third box?
How many total combinations?
27
Example 38.1
How many different ways can the numbers 3, 5, 7, and 8 be
arranged in order if no repetition is permitted?
4 items – 4 boxes
4 3 2 1
How many choices for the first box?
Third box?
Total combinations?
Fourth box?
24
Second box?
Example 38.2
How many 4-letter signs can be made from the letters in the
word EQUAL if repetition is permitted?
4 items – 4 boxes
5 5 5 5
How many choices for the first box?
Third box?
Total combinations?
Fourth box?
625
Second box?
Example 38.3
A multiple choice test has 8 questions, and there are 4
possible choices to each question. How many different sets of
answers are possible?
8 items – 8 boxes
4 4 4 4 4 4 4 4
How many choices for the first box?
Third box?
Fifth box?
Fourth box?
Sixth box?
Seventh box?
Eighth box?
Total combinations?
65536
Second box?
Example 38.4
How many 3-letter signs can be made from the letters in the
word NUMERAL if no repetition is permitted?
3 items – 3 boxes
7 6 5
How many choices for the first box?
Third box?
Total combinations?
210
Second box?
Designated Roots
The Zero Factor Theorem – If the product of two numbers equals
zero, then one (or more) of the quantities must equal zero.
Sample:
(x – 4)(x + 7) = 0
Either x – 4 = 0 or x + 7 = 0, so x = 4 or x = -7.
x2 + 3x – 28 = 0 has the same roots as 2x2 + 6x – 56 = 0.
Factor out a two from the second.
Example 38.5
Write the quadratic equation with a lead coefficient of 1
whose roots are −
3
1
and
.
5
2
If these are the roots then they came from:
3

x
+


5


and
1

x
−


2


Put these together to get the factored equation of:
3 
1

 x +  x −  = 0
5 
2

Multiply
x2 +
1
3
x− =0
10
10
Lead coefficient = 1, since
there is 1 x2.
Example 38.6
Write the quadratic equation with a lead coefficient of 1
whose roots are 1+ 2 and 1− 2 .
If these are the roots then they came from:
(x − (1+ 2 ))
( (
and x − 1− 2
))
Put these together to get the factored equation of:
(x − (1 + 2 ))(x − (1 − 2 )) = 0
Simplify first.
(x − 1 − 2 )(x − 1 + 2 ) = 0
Multiply
x2 − x + 2x − x +1− 2 − 2x + 2 − 2 = 0
x2 − 2x −1 = 0
Lead coefficient = 1, since
there is 1 x2.
Overall Average Rate
Overall average rate is total accomplishments divided by total time.
Common sense sample.
Travel at 60 mph for 59 minutes and 100 mph for 1 minute
What is the overall average rate for the trip?
Not 80 mph.
Maybe 61 mph or so.
Example 38.7
Frank and Judy drive for 100 miles at 50 mph. They then drive 180
miles at 60 mph. What is Frank and Judy’s overall average rate over
the whole trip?
2
How many legs to the trip?
Overall Distance
Overall Average Rate =
Overall Time
Overall Distance = 100 + 180
Overall Time
Leg 1
t=
d
v
=
100
50
= 280 miles
Find the time for each leg and combine.
= 2 hrs
Overall Time = 2 + 3
Leg 2
t=
= 5 hrs
Overall Average Rate =
280
= 56 mph
5
d
v
=
180
60
= 3 hrs
Example 38.8
Frank and Judy drive for 100 miles at 50 mph. They then drive 180
miles at 60 mph. How fast must they drive for the last 120 miles to
have an overall average rate of 60 mph for the entire trip?
How many legs to the trip?
3
Overall Distance
Overall Average Rate =
Overall Time
Overall Distance = 100 + 180 + 120 = 400 miles
Overall Time
Leg 1
Leg 3
180
d
=
= 3 hrs
v
60
Overall Distance
2
400
Overall Time =
= 6 hrs
=
Overall Average Rate
3
60
t=
v=
d
v
=
100
50
Find the time for each leg and combine.
= 2 hrs
d
120
= 72 mph
=
2
t
6 −5
3
Leg 2
t=
Practice
a) The average speed for the first 300 miles was 60 mph and for the
next 200 miles was 20 mph. What should be the speed for the last 500
miles so that the overall average speed for the trip would be 40 mph?
How many legs to the trip?
3
Overall Distance
Overall Average Rate =
Overall Time
Overall Distance = 300 + 200 + 500 = 1000 miles
Overall Time
Leg 1
Leg 3
Find the time for each leg and combine.
200
d
=
= 10 hrs
v
20
Overall Distance
1000
Overall Time =
= 25 hrs
=
Overall Average Rate
40
t=
v=
d
v
=
300
60
= 5 hrs
500
d
=
= 50 mph
t 25 − 15
Leg 2
t=
b) How many 5-letter signs can be made from the letters in the word
LEFTY if repetition is permitted?
5 items – 5 boxes
5 5 5 5 5
How many choices for the first box?
Third box?
Total combinations?
Fourth box?
3125
Second box?
Fifth box?
c) Write the quadratic equation with a lead coefficient of 1
2
4
whose roots are −
and
.
5
3
If these are the roots then they came from:
2

x+ 
5

and
4

x− 
3

Put these together to get the factored equation of:
2 
4

+
−
x
x


=0
5 
3

Multiply
4
2
8
x − x+ x− =0
3
5
15
2
Lead coefficient = 1, since
there is 1 x2.
14
8
x − x− =0
15
15
2

 k2

− k 
d) Evaluate: ∑ 
k = −1 4

3

  (3)2
  (2 )2
  (1)2
 (− 1)2
  (0 )2









− (3)
− (2 ) + 
− (1) + 
− (0 ) + 
=
− (− 1) + 

  4
  4
  4
 4
  4
 5  3
 3
=   +  −  + (− 1) +  − 
 4  4
 4
 5   10 
=   + − 
 4  4 
=−
5
4