Transcript Pascal`s Triangle

Blaise Pascal
Born: June 19, 1623
Clermont Auvergne, France
Died: 19 Aug 1662
Paris, France
Pascal’s Triangle
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1
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1
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1 7
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5
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15
21
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1
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1
2
6
10
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10
20
35
1
5
15
35
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6
21
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7
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Creating Pascal’s Triangle
1
(x + y)0 = 1
1
(x + y)1 = 1x + 1y
1
(x + y)2 = 1x 2 + 2xy + 1y 2
1
(x + y)3 = 1x 3 + 3x2y + 3xy 2 + 1y3
(x + y)4 = 1x 4 + 4x3y + 6x2y 2 + 4xy3 + 1y4
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1
2
3
4
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1
3
6
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Answer the following questions before going to the next slide:
1) Where do the numbers come from to form the triangle?
2) How do you continue the triangle for three more rows?
1) Where do the numbers come from to form the triangle?
The entries in the triangle are the coefficients of the
terms of the binomial expression – (x + y)n
1
(x + y)0 = 1
1
(x + y)1 = 1x +1 y
1
(x + y)2 = 1x 2 + 2xy + 1y 2
1
(x + y)3 =1x 3 + 3x2y + 3xy 2 + 1y3
2
3
4
1
(x + y)4 =1x 4 + 4x3y + 6x2y 2 + 4xy3 + 1y4
1
1
1
3
6
4
2) How do you continue the triangle for three more rows?
In the triangle the first and last terms are always 1. The terms in
between are the sum of two adjacent terms. (watch above)..
The next three rows would be:
1
1
1
15
6
7
21
10
10
5
35
20
35
1
5
15
21
6
1
7
1
1
Patterns in Pascal’s Triangle
Look at the row in the blue ellipse.
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1
1
1
1 7
3
5
6
2
15
21
4
10
20
35
6 +10 =16
1
6
10
3+6=9
1
3
4
1
Example: 1 + 3 = 4
1
1
1
Add the numbers two at a time.
What kind of answers do you get?
1
5
15
35
1
6
21
The answers are all
perfect squares.
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7
1
Try to find some other number patterns.
Patterns in Pascal’s Triangle
Look at the part of a diagonal in the rectangle.
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1
1
1
3
5
6
2
15
1 7 21
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6
10
4
10
20
35
Where do you find the answer in Pascal’s
Triangle?
1
3
4
1
1 + 3 + 6 + 10 = 20
1
1
1
Add the three numbers together.
1
5
15
35
Some people call this the
Hockey Stick Pattern.
1
6
21 7
Try this for other parts of diagonals?
1
1
Combinations Using Formulas
1. Suppose you have three socks and want to figure out how
many different ways you can choose two of them to wear. You
don’t care which feet you put them on, it only matters which
two socks you pick. This problem amounts to the question
“how may different ways can you choose two objects from a set
of three objects?”
Suppose you have S1, S2, S3
(sock 1, sock 2, sock 3)
You could choose: S1,S2 S1,S3 S2,S3 (3 ways to choose 2 socks)
Formula:
3C2
n Cr
=
n!
(n  r)! r!
= 3! / (2!1!) = (3*2*1) / (2*1*1) = 6/2 = 3
Combinations Using Formulas
(continued)
Suppose you wanted to choose one sock.
3C1
S1 or S2 or S3
(3 ways)
= 3! / 1!2! = 3 ways
Suppose you wanted to choose all three socks {S1, S2, S3} (1 way)
3C3
= 3! / 3!0! = 1 way
Combinations and Pascal’s
Triangle
Now use Pascal’s Triangle to determine how many ways you can
choose 1 sock, 2 socks, or 3 socks.
1
1 1
1 way for choosing
1 2 1
three socks
1 3 3 1
3 ways for
choosing one sock
3 ways for
choosing two socks
1) Ignore the 1’s running down the left-hand side of Pascal’s Triangle.
2) Choose the row where 3 is in the second position because you have 3 socks.
n!
Compare using the previous formula method. nCr =
(n  r)! r!
A basketball coach is criticized in the newspaper
for not trying out every combination of players. If
the team roster has 10 players, how many fiveplayer combinations are possible?
Use can use the combination formula to find how
many combinations of five players are possible.
10C5 =
10!
= 252 ways to choose five players
5!5!
Or…
Use Pascal’s Triangle to find the number of ways.
You choose the row with 10 after the 1 because
there are 10 players from which to choose.
The row where 10 is the first number after the 1 is:
1
10
45
120
210
252
210
120
45
10
1
This row tells you there are:
10 ways to choose one player.
210 ways to choose four players.
45 ways to choose two players.
252 ways to choose five players.
120 ways to choose three players.
10C5 =
10! = 252
5!5!
Pascal’s Triangle and the formula
result in the same answer.
There are 8 people in an office. 3 are to
be chosen for the grievance committee. In
how many ways can 3 people be chosen
out of the eight in the office?
You can use the combination formula to find how many
combinations of three people are possible.
8!
= 56 ways to choose five players
8C3 =
5!3!
OR …
Use Pascal’s Triangle to find the number of ways.
You choose the first row that has an 8 after the
one since there are 8 people to choose from.
The row where 8 (you have 8 people) is the first number after the 1:
1 8 28 56 70 56 28 8 1
This row of numbers tells you:
8 ways to choose one person.
28 ways to choose two people.
56 ways to choose three people.
8C3 =
8!
= 56
5!3!
Pascal’s triangle gives the
same answer as the formula.
Fun with Pascal’s Triangle
Suppose your name is Tom
Arrange your name in Pascal’s Triangle form with one letter per line.
Find how many different paths you can take to spell your name.
T
O
1
O
1
M
M
M
1
2
1
1 path ends with the M on the left
2 paths end with the M in the middle
1 path ends with the M on the right.
There are 4 paths:
black, red; black purple;
blue orange; blue, green
1
1
2
1
The name arrangement is
like Pascal’s Triangle.
The number of letters in
your name determines how
many rows you use.
The sum of the numbers
in the last row tells you how
many paths will spell your
name.
I’ll try my name. You try your name.
Suppose I only want to count the
part that is not covered by the
triangles?
G
R
O
N
B
E
R
G
1
G
7
O
E
21
N
B
E
R
G
O
N
B
R
R
N
B
E
R
How many ways will there be?
B
E
R
E
R
R
G
G
G
G
G
35
35
21
7
1
How many paths spell Gronberg?
128 ways
70 ways
Fun and Games
Use the website to
learn how to make
a “pinball”
machine, as in the
picture, based on
Pascal’s Triangle.
http://www.math.bcit.ca/entertainment/pascaltr/index.shtml
The Pinball Game
The gray lines indicate the
paths. There are 5 bins
where the marble can finish.
What is the probability that the
ball will end up in any given bin?
There are 16 paths (sum of the number
of paths to each bin) and the ball is
equally likely to follow any of the paths.
•Probability = 1/16 to end in the 1st
(leftmost) bin.
•Probability = 4/16 to end in the 2nd bin.
•Probability = 6/16 to end in 3rd bin.
•Probability = 4/16 to end in 4th bin.
•Probability = 1/16 to end in the 5th bin.
1 4
6
4
1
Number of path ending in each bin. The 5th row
of Pascal’s Triangle because there are 5 bins.