Definition of the Factorial
We define n! recursively by
0! = 0, 1! = 1,
n! = n(n - 1)!
5! = 5(4)(3)(2) = 120
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!
If we want to select only three letters from the word "Cindy" then we have
(5)(4)(3) = 5!/(5 -
The number of permutations of n distinct objects
taken r at a time is
nPr = n!/(n - r)!
You can find this button on the TI 85 calculator by hitting Math -> Prob
How many ways are there of scrambling the name Tamara Heether?
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
We must divide by 2!3!2!3! to get
13!/[2! 3! 2! 3!] = 43,243,200
If there are n objects with n1 duplicates
of one kind, n2 duplicates of a second kind, ...,
duplicates of a kth kind, then the number of distinguishable permutations
of these n objects is
How many ways are there to scramble your first and
How many different five card poker hands are there?
First note that there are 52P5 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
52P5/5! = 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.
The number of ways of choosing r objects from
order does not matter is
= n!/(n - r)!r!
The Binomial Theorem
(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?
How many ways are there to select two x's from the five?
investigations lead us to believe that
(x + y)5 = 5C5 x5 +
x4y + 5C3 x3y2 + 5C2
x2y3 + 5C1 xy4 5C0 y5
(x + y)n = Si = 0n
nCn - i
(3x - 2y)4
The formula gives us
5C4 (3x)4 +
5C3 (3x)3(-2y) + 5C2
(3 x)2(-2y)2 + 5C1 (3 x)(-2y)3
+ 10(27x3)(-2y) + 10(9x2)(4y2) + 5(3x)(-8y3)
= 405x4 - 540x3y
+ 360x2y2 - 120xy3 + 16y4