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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
