Pascal’s Triangle and the Binomial Theorem

Chapter 5.2 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

Pascal’s Triangle outer values: always 1

inner values: add the two values diagonally above

Pascal’s Triangle – the counting shortcut

1 1 1

1 2 1 1 3 3 1

1 4 6 4 1 1 5 10 10 5 1

1 6 15 20 15 6 1

sum of each row is a power of 2 1 = 20

2 = 21

4 = 22

8 = 23

16 = 24

32 = 25

64 = 26

Pascal’s Triangle – Why? Combinations!

e.g. choose 2 items from 5 go to the 5th row, the 2nd number = 10 (always start

counting at 0) Binomial Theorem Patterns Physical applications modeling the electrons in each shell of an atom (google

‘Pascal’s Triangle electron’)

Pascal’s Triangle – Cool Stuff 1

1 1 1 2 1

1 3 3 1 1 4 6 4 1

1 5 10 10 5 1 1 6 15 20 15 6 1

each diagonal is summed below and to the left

called the “hockey stick” property

Pascal’s Triangle – Cool Stuff e.g., numbers

divisible by 5 similar

patterns for other numbers

http://www.shodor.org/interactivate/activities/pascal1/

Pascal’s Triangle can also be seen in terms of combinations n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

0

0

0

1

2

4

3

4

4

4

0

5

1

5

2

5

3

5

1

1

0

2

1

2

2

2

0

3

1

3

2

3

3

3

0

4

1

4

4

5

5

5

0

6

1

6

2

6

3

6

4

6

5

6

6

6

Pascal’s Triangle - Summary symmetrical down the middle outside number is always 1 second diagonal values match the row

numbers sum of each row is a power of 2

sum of nth row is 2n

Begin count at 0 number inside a row is the sum of the two

numbers above it

The Binomial Theorem

the term (a + b) can be expanded: (a + b)0 = 1 (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Blaise Pascal (for whom the Pascal computer language is named) noted that there are patterns of expansion, and from this he developed what we now know as Pascal’s Triangle. He also invented the second mechanical calculator.

So what does this have to do with the Binomial Theorem?

remember that the binomial expansion: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

and the triangle’s 4th row is 1 4 6 4 1 Pascal’s Triangle allows you to determine the

coefficients The exponents on the variables form a

predictable pattern The exponents of each term sum to n

The Binomial Theorem

nrrnnnn

n

bn

nba

r

nba

nba

na

n

ba

......210

)(

221

rrnn bar

nba

is term t the)( binomial for the so 1r

A Binomial Expansion

Expand (x + y)4

432234

444334224114004

4

464

4

4

3

4

2

4

1

4

0

4

yxyyxyxx

yxyxyxyxyx

yx

Another Binomial Expansion

Expand (a – 4)5

1024128064016020

)1024()1()256()5()64()10()16()10()4()5()1()1(

)4(5

5)4(

4

5)4(

3

5)4(

2

5)4(

1

5)4(

0

5

4

2345

012345

555445335225115005

5

aaaaa

aaaaaa

aaaaaa

a

Some Binomial Examples

what is the 6th term in (a + b)9? don’t forget that when you find the 6th term, r = 5

what is the 11th term of (2x + 4)12

54559 1265

9baba

22101012 2768240641048576)4(664)2(10

12xxx

Look at the triangle in a different way r0 r1 r2 r3 r4 r5 n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 n = 6 1 6 15 20 15 6 1

for a binomial expansion of

(a + b)5, the term for r = 3 has a coefficient of 10

And one more thing… remember that for the inner numbers in the

triangle, any number is the sum of the two numbers above it

for example 4 + 6 = 10 this suggests the following:

which is an example of Pascal’s Identity

2

5

2

4

1

4

1

1

1 r

n

r

n

r

n

r

n

1r

n

1

1

r

n

For Example…

6

9

6

8

5

8

4

7

4

6

3

6

How can this help us solve our original problem?

1 5 15 35 70 126

1 4 10 20 35 56

1 3 6 10 15 21

1 2 3 4 5 6

1 1 1 1 1

so by overlaying Pascal’s Triangle over the grid we can see that there are 126 ways to move from one corner to another

How many routes pass through the green square? to get to the green

square, there are C(4,2) ways (6 ways)

to get to the end from the green square there are C(5,3) ways (10 ways)

in total there are 60 ways

How many routes do not pass through the green square? there are 60 ways

that pass through the green square

there are C(9,5) or 126 ways in total

then there must be 126 – 60 = 66 paths that do not pass through the green square

MSIP/ Homework

Read the examples on pages 281-287, in particular the example starting on the bottom of page 287 is important

Complete p. 289 #1, 2aceg, 3, 4, 5, 6, 8, 9, 11, 13