Algebra III
Lesson 28
Factorial Notation – Abstract Rate Problems
Factorial Notation
x! Î read as ‘x factorial’
4! = 4 • 3 • 2 • 1
12! = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
The factorial for a given number means to multiply all the
integer numbers from the given number to one.
Example 28.1
Evaluate 5!
Rewrite
5! = 5 • 4• 3 • 2 • 1
5 • 4 = 20 • 3 = 60 • 2 = 120 •1 = 120
5! =120
Example 28.2
Evaluate
9!
without using a calculator.
3!5!
9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
=
3!5!
3 ⋅ 2 ⋅1 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
Reduce
=
9 ⋅8⋅ 7 ⋅ 6
3 ⋅ 2 ⋅1
Reduce more
=9•8•7
= 72 • 7
= 504
Note: 20! = 20 • 19 • 18 • 17 • 16 • 15!
20! = 20 • 19!
Example 28.3
Evaluate
14!
without using a calculator.
6!11!
Since there are two factorials in the bottom rewrite 14! in
terms of the biggest one as in the note).
14! 14 ⋅13 ⋅12 ⋅11!
=
6!11!
6!11!
Reduce the 11!
14 ⋅13 ⋅12
=
6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
Reduce more
=
7 ⋅13
6⋅5
=
91
30
Example 28.4
Use the factorial key on a calculator to help evaluate 11 • 10 • 9 • 8.
This is the start of 11!, but from 7 down is missing.
Let’s put them back.
Since we just can’t throw 7! into a problem, we
have to get them there legally.
Multiply by 1 or 7!/7!.
11 ⋅10 ⋅ 9 ⋅ 8 ⋅
=
11!
7!
= 7920
7!
7!
Abstract Rate Problems
Everything is in terms of letters, numbers are few and far between.
You end up with a relationship between pieces or concepts,
like distance, rate, or time.
As in all problems, watch your units, make sure they agree.
All feet, all kph, all minutes, etc.
Remember the basic formulas.
d = rt
r=
d
t
t=
d
r
These problems tend to have to be done in two parts.
1st – You will be given 2 of the 3 pieces, and find
the 3rd using this information.
2nd – Some things are changed and you are again
told about 2 of the 3 pieces, and have to find the
missing piece.
Example 28.5
The train traveled m miles at p miles per hour and still arrived 1
hour late. How fast should the train have traveled to have
arrived on time?
Original Data
d =m
r=p
d m
t= =
r p
r = ??
t=
New Data
d =m
Find r.
r=
d
m
=
t m −1
p
m
−1
p
Multiply by p top and bottom to clear the complex fraction.
m
p
r=
•
m
−1 p
p
r=
mp mi
m − p hr
Example 28.6
The boat traveled k miles in t hours and was 40 miles short of
the goal when the gun went off. If the skipper tried again, how
long would it taker to reach the goal if she increased the rate of
travel by 10 mph?
Original Data
d =k
r=
d k
=
t t
t =t
r=
k
+ 10
t
t = ??
New Data
d = k + 40
Find t.
t=
d k + 40
=
r k + 10
t
Multiply by t top and bottom to clear the complex fraction.
k + 40 t
t=
•
k
+ 10 t
t
(
k + 40 )t
t=
hrs
k + 10t
Practice
a) The bus traveled x miles at p miles per hour and still arrived
2 hours late. How fast should the bus have traveled to have
arrived on time?
Original Data
d=x
r=p
d x
t= =
r p
r = ??
t=
New Data
d=x
r=
Find r.
d
x
=
x
t
−2
p
x
−2
p
Multiply by p top and bottom to clear the complex fraction.
x
p
r=
•
x
−2 p
p
r=
xp mi
x − 2 p hr
b) Redig ran for T hours at R miles per hour, but ended up 20
miles short of the goal. If he tried again and increased his
speed by 5 miles per hour, how long would it taker to reach the
goal?
Original Data
d = rt = RT
r=R
t =T
r = R+5
t = ??
New Data
d = RT + 20
d RT + 20
Find t. t = =
r
R+5
8!
c) Evaluate
2!4!
without using a calculator.
8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4!
=
2!4!
2 ⋅1 ⋅ 4!
8⋅7 ⋅6⋅5
=
2 ⋅1
=4•7•6•5
= 28 • 30
= 840
9!
d) Evaluate
3!3!
without using a calculator.
9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3!
=
3!3!
3 ⋅ 2 ⋅1 ⋅ 3!
=
9 ⋅8⋅ 7 ⋅ 6 ⋅5⋅ 4
3 ⋅ 2 ⋅1
=9•8•7•5•4
= 72 • 7 • 20
= 504 • 20
= 10080
e) Draw the necessary reference triangle and evaluate:2 cos(− 330°)
1) Sketch
4) Replace
90°
2 cos(− 330°) = 2(+ cos 30°)
30°
180°
-330°
0°
360°
270°
2) Related angle
30°
3) Sign
cos is like x, x would be +, so
cos is +.
5) Solve
= 2(cos 30°)
 3

= 2

 2 
= 3
2
60°
30°
3
1