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Pascal’s triangle

Math 40210, Fall 2012

October 25, 2012

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 1 / 21

Pascal’s triangle - symbolic

(00

)(10

) (11

)(20

) (21

) (22

)(30

) (31

) (32

) (33

)(40

) (41

) (42

) (43

) (44

)(50

) (51

) (52

) (53

) (54

) (55

)(60

) (61

) (62

) (63

) (64

) (65

) (66

)(70

) (71

) (72

) (73

) (74

) (75

) (76

) (77

)(80

) (81

) (82

) (83

) (84

) (85

) (86

) (87

) (88

). . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 2 / 21

Pascal’s triangle - filling out the edges

11 1

1(2

1

)1

1(3

1

) (32

)1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 3 / 21

Pascal’s triangle - using formula

11 1

1(2

1

)1

1(3

1

) (32

)1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 4 / 21

Pascal’s triangle - using formula

11 1

1 1 + 1 11(3

1

) (32

)1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 5 / 21

Pascal’s triangle - using Pascal’s formula

11 1

1 2 11(3

1

) (32

)1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 6 / 21

Pascal’s triangle - using Pascal’s formula

11 1

1 2 11(3

1

) (32

)1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 7 / 21

Pascal’s triangle - using formula

11 1

1 2 11 1 + 2 2 + 1 11(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 8 / 21

Pascal’s triangle - using formula

11 1

1 2 11 3 3 1

1(4

1

) (42

) (43

)1

1(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 9 / 21

Pascal’s triangle - using formula

11 1

1 2 11 3 3 1

1 4 6 4 11(5

1

) (52

) (53

) (54

)1

1(6

1

) (62

) (63

) (64

) (65

)1

1(7

1

) (72

) (73

) (74

) (75

) (76

)1

1(8

1

) (82

) (83

) (84

) (85

) (86

) (87

)1

. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 10 / 21

Pascal’s triangle - using formula

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 11 / 21

Pascal’s triangle - the row sums

1 = 11 + 1 = 2

1 + 2 + 1 = 41 + 3 + 3 + 1 = 8

1 + 4 + 6 + 4 + 1 = 161 + 5 + 10 + 10 + 5 + 1 = 32

1 + 6 + 15 + 20 + 15 + 6 + 1 = 641 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128

1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 12 / 21

Pascal’s triangle - the symbolic row sums(00

)= 1(1

0

)+(1

1

)= 2(2

0

)+(2

1

)+(2

2

)= 4(3

0

)+(3

1

)+(3

2

)+(3

3

)= 8(4

0

)+(4

1

)+(4

2

)+(4

3

)+(4

4

)= 16(5

0

)+(5

1

)+(5

2

)+(5

3

)+(5

4

)+(5

5

)= 32(6

0

)+(6

1

)+(6

2

)+(6

3

)+(6

4

)+(6

5

)+(6

6

)= 64(7

0

)+(7

1

)+(7

2

)+(7

3

)+(7

4

)+(7

5

)+(7

6

)+(7

7

)= 128(8

0

)+(8

1

)+(8

2

)+(8

3

)+(8

4

)+(8

5

)+(8

6

)+(8

7

)+(8

8

)= 256

. . .

Identity: for all n ≥ 0,n∑

i=0

(ni

)= 2n

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 13 / 21

Pascal’s triangle - the alternating row sums

1 = 11 − 1 = 0

1 − 2 + 1 = 01 − 3 + 3 − 1 = 0

1 − 4 + 6 − 4 + 1 = 01 − 5 + 10 − 10 + 5 − 1 = 0

1 − 6 + 15 − 20 + 15 − 6 + 1 = 01 − 7 + 21 − 35 + 35 − 21 + 7 − 1 = 0

1 − 8 + 28 − 56 + 70 − 56 + 28 − 8 + 1 = 0. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 14 / 21

Pascal’s triangle - the symbolic alternating row sums(00

)= 1(1

0

)−(1

1

)= 0(2

0

)−(2

1

)+(2

2

)= 0(3

0

)−(3

1

)+(3

2

)−(3

3

)= 0(4

0

)−(4

1

)+(4

2

)−(4

3

)+(4

4

)= 0(5

0

)−(5

1

)+(5

2

)−(5

3

)+(5

4

)−(5

5

)= 0(6

0

)−(6

1

)+(6

2

)−(6

3

)+(6

4

)−(6

5

)+(6

6

)= 0(7

0

)−(7

1

)+(7

2

)−(7

3

)+(7

4

)−(7

5

)+(7

6

)−(7

7

)= 0(8

0

)−(8

1

)+(8

2

)−(8

3

)+(8

4

)−(8

5

)+(8

6

)−(8

7

)+(8

8

)= 0

. . .

Identity: for all n ≥ 1,n∑

i=0

(−1)i(

ni

)= 0

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 15 / 21

Pascal’s triangle - summing along diagonals

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 16 / 21

Pascal’s triangle - summing along diagonals

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 17 / 21

Pascal’s triangle - summing along diagonals

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 18 / 21

Pascal’s triangle - summing along diagonals

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . .

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 19 / 21

Pascal’s triangle - symbolic sum along diagonals(00

)(10

) (11

)(20

) (21

) (22

)(30

) (31

) (32

) (33

)(40

) (41

) (42

) (43

) (44

)(50

) (51

) (52

) (53

) (54

) (55

)(60

) (61

) (62

) (63

) (64

) (65

) (66

)(70

) (71

) (72

) (73

) (74

) (75

) (76

) (77

)(80

) (81

) (82

) (83

) (84

) (85

) (86

) (87

) (88

). . .

Identity: for all k ≥ 0, and s ≥ ks∑

i=k

(ik

)=

(s + 1k + 1

)Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 20 / 21

The vandermonde convolutionEach week, IN lottery selects m numbers from a bag with m + nnumbers. When you buy a lottery ticket, you select ` numbersfrom the m + n. You have a k-win if you have exactly k of the theselected numbers among your `.

How many tickets are k -wins?

It’s(m

k

)( n`−k

)(automatically 0 if k > ` or k > m)

How many tickets in all?

(m + n

`

)=∑

k

(mk

)(n

`− k

) ormin{m,`}∑

k=0

(mk

)(n

`− k

)

Math 40210 (Fall 2012) Pascal’s triangle October 25, 2012 21 / 21