
Definition of the Factorial
We define n! recursively by
0! = 0, 1! = 1,
n! = n(n  1)!
Example:
5! = 5(4)(3)(2) = 120
Example:
Suppose that we are interested in how many ways there are in scrambling the
letters of the name "Cindy". We have 5 choices for the first letter,
once we have chosen the first letter there are 4 choices for the second letter,
and then three choices for the third letter, two for the fourth letter, and
only one choice for the last letter. Hence there are
5(4)(3)(2)(1) = 5!
choices.
Permutations
Example
If we want to select only three letters from the word "Cindy" then we have
(5)(4)(3) = 5!/(5 
3)!
choices.
Definition
The number of permutations of n distinct objects
taken r at a time is
_{n}P_{r} = n!/(n  r)! 
You can find this button on the TI 85 calculator by hitting Math > Prob

Distinguishable Permutations
Example
How many ways are there of scrambling the name Tamara Heether?
Solution:
If there were no duplicate letters the solution would
be 13!, but this is not the case. There are
2 T's, 3 A's
2 R's and
3E's
We must divide by 2!3!2!3! to get
13!/[2! 3! 2! 3!] = 43,243,200
Theorem
If there are n objects with n_{1} duplicates
of one kind, n_{2} duplicates of a second kind, ...,
n_{k}
duplicates of a kth kind, then the number of distinguishable permutations
of these n objects is
n!/(n_{1}!n_{2}!...n_{k}!) 
Exercise:
How many ways are there to scramble your first and
last name?

Combinations
Example
How many different five card poker hands are there?
Solution
First note that there are _{52}P_{5} different ordered five card poker hands,
however, two hands that have the same five cards, but in a different order should not
be counted as distinct hands. Since there are 5! ways of ordering five
cards, we have
_{52}P_{5}/5! = 52!/[5!(52  5)!] = 2,598,960
different poker hands.
Note that only four of these hands are Royal Flushes, hence there is a 4
in 2,598,960 or about one in half a million chance of receiving a Royal
Flush in a 5 card stud poker game.
Theorem
The number of ways of choosing r objects from
n where
order does not matter is
_{n}C_{r}
= n!/(n  r)!r! 

The Binomial Theorem
consider
(x + y)^{5} = (x + y)(x + y)(x + y)(x + y)(x + y)
Q: How many ways are there to select all x's?
A: 1 way.
Q: How many ways are there to select 4x's from the 5 possible?
A: _{5}C_{4}^{
}ways
Exercise:
How many ways are there to select two x's from the five?
These
investigations lead us to believe that
(x + y)^{5} = _{5}C_{5} x^{5} +
_{5}C_{4}
x^{4}y + _{5}C_{3} x^{3}y^{2} + _{5}C_{2}
x^{2}y^{3} + _{5}C_{1} xy^{4} _{5}C_{0} y^{5}
Theorem
(x + y)^{n} = S_{i = 0}^{n}
_{n}C_{n  i}
x^{ni}y^{i}

Example
Find
(3x  2y)^{4}
Solution
The formula gives us
_{5}C_{4} (3x)^{4} +
_{5}C_{3} (3x)^{3}(2y) + _{5}C_{2}
(3 x)^{2}(2y)^{2} + _{5}C_{1} (3 x)(2y)^{3}
_{5}C_{0} (2y)^{4}
= 5(3^{4}x^{4})
+ 10(27x^{3})(2y) + 10(9x^{2})(4y^{2}) + 5(3x)(8y^{3})
+ (16y^{4})
= 405x^{4}  540x^{3}y
+ 360x^{2}y^{2}  120xy^{3} + 16y^{4}