scribe post: counting

Hello guys... This is Richard and it's my third scribe post. But before we proceed to our lesson for today, let's take a look to the notes on last thursday (November 16) since there's no scribe that day.

COUNTING (a.k.a. COMBINATORICS)

The Fundamental Principle of Counting

If there are m ways to do the first thing and n ways to do a second thing then there are m • n ways to do both things.

example:

How many different outfits can be made if a person has 3 pairs of pants and 4 shirts?

solution:

3 ways to choose pants

4 ways to choose shirts


۞ 3 x 4 = 12 ways to choose both

Factorial Notation

Definition:

n! = n (n-1) (n-2) (n-3) . . . 3, 2, 1

0! = 1

examples:

4! = 4 • 3 • 2 • 1

= 24

6! = 6 • 5 • 4 • 3 • 2 • 1

= 720

Note: FACTORIAL NOTATION IS UNDEFINED FOR NEGATIVE NUMBERS

n! = n (n-1)!

Permutation

An order arrangement of objects.

FORMULA

nPr = n!/ (n-r)!

n is the number of objects to pick from

r is the number of object to be arrange

nPr is read as " n Pick r"

nPr given a set of n objects. How many orderd arrangements can be made using only r of them at a time.

examples:

There are 8 horses in a race. In how many different ways can they finish 1st, 2nd, 3rd.

solution:

(by fundamental principle of counting)

8 • 7 • 6

- - - = 336

1 2 3

(by Permutation)

8P3 = 8! / (8-3)!

= 8!/ 5!

= 336

Permutation of Non-distinguishable objects

The number of ways to arrange "n" objects that contain K1, K2, K3. . . .etc. Set of non-distinguishable objects is given by:

n!

__________

K1! K2! K3!...

example:

How many different words can be made from the letters in the words:

(a) book (b) Mississippi

(a) # letters = 4

# of O's = 2

= 4! / 2!

= 12

(b) # letters = 11

# of I's = 4

# of S's = 4

# of P's = 2

= 11! / (2! 4! 4!)

November 21, 2006

In the beginning of the class, Mr.K give us the following problems without using a calculator.

Simplify

(a) 8! / 4!

8•7•6•5•4!

_______ 4! is reduce

4!

so we left with 8 • 7 • 6 • 5

= 1680

(b) 7! - 5!

7 • 6 • 5! - 5!

5! (7 • 6 - 1) factoring by 5!

120 (41) = 4920

(c) 10! 4! / 8! 6!

10•9•8! 4!

_______ 8! & 4! are reduce

8! 6•5•4!

10•9

____ = 3

6•5

Write using factorial

(a) 5•4•3•2•1

= 5!

(b) 5•4•3

= 5! / 2!

(c) 20•19•18•17

=20! / 16!

Simplify

(a) (n+1)! / n!

(n+1) n!

______ from the definition of factorial notation n! = n ( n-1)!

n!

= n + 1

(b) (n-1)! / n!

(n-1)!

_____

n (n-1)!

= 1 / n

(c) (n+1) (n-1) / (n!)²

(n+1) n! (n-1)!

__________

n! n (n-1)!

= (n+1) / n

(d)

(2n-1)! (n+1)!

__________

(2n+1)! (n-1)!

(2n-1)! (n+1)( n ) (n-1)!

_________________

(2n+)(2n)(2n-1)! (n-1)!

n(n+1)

_______ n is reduce

2n(2n+1)

n+1

____

4n+2

SOLVE:

(a) 3! (n-1)!

_______ = 72

(n-3)!

(n-1) ! 72

____ = __

(n-3)! 6

(n-1)(n-2) = 12 (n-3)! is reduce.

n² -3n +2 = 12

n²-3n-10 = 0

(n-5) (n+2) = 0

n= 5 n= -2

-2 is rejected because the factorial of negative numbers is undefined.

Always indicate if it is accepted or reject.

(b) (2n-1)!

________ = 10

2! (2n -3)

(2n-1) (2n-2) (2n-3)

______________ = 20

(2n-3)

(2n-1) (2n-2) = 20

4n² -6n - 18 = 0

2n² -3n - 9 = 0 it is divided by 2

(2n+3) (n-3)

n= -3/2 n = 3

-3/2 is rejected

I'm sorry guys if this is not nice..I'll fix it later. I just can't upload the images.

By the way the next scribe will be John.