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Transcript
Patterns and Combinatorics
Pascal’s Triangle
Pascal’s Triangle: click to see movie
Triangular numbers
Triangular numbers: short formula:
Proof:
Two triangles of side n form a rectangle of sides n and n+1:
Here n=4:
For each polynomial, connect all pairs of vertices by lines. Some
of these lines will be the sides, some will be diagonals.
How many such lines are there in total in each polynomial?
1
2
1
2
1
2
2
3
3
4
1
2
2
7
3
5
3
4
5
1
2
7
6
6
4
1
8
5
3
4
3
6
4
5
1
9
3
8
4
7
5
6
We counted all lines connecting the vertices of the polygons:
Can you explain the pattern?
1
2
1
1
2
5
2
3
2
3 lines
3
4
6 lines
+3
+4
2
1
3
6
4
3
1
4
10 lines
5
15 lines
+5
1
1
9
2
2
7
3
8
3
3
6
4
+6
21 lines
7
4
5
5
+7
28 lines
8
4
6
7
5
+8
6
36 lines
The number of lines connecting the vertices of a polygon with n
sides is the (n-1)th triangular number:
2
The n-th vertex
contributes
n-1
red lines:
1
3
4
also denoted
counts how many pairs of 2 vertices we can
form from n given vertices.
Each pair determines a line.
Count the rows in Pascal’s triangle starting from 0.
The entry on the n-th horizontal row, and k-th slanted row in Pascal’s triangle:
= number of ways to choose a set of k
objects from among n given objects.
1=
=
1=
1=
1=
1=
5th horizontal row
1=
3=
3=
6=
10=
2nd slanted row
1=
2=
4=
5=
1=
1=
4=
10=
1=
5=
1=
Any three vertices of a polygon form a triangle.
How many such triangles are there in each polynomial?
1
2
1
2
1
2
2
3
3
4
1
2
2
7
3
5
3
4
5
1
2
7
6
6
4
1
8
5
3
4
3
6
4
5
1
9
3
8
4
7
5
6
Triangles in each polynomial:
1
2
1
1
2
5
2
3
2
1=
3
4
4=
3=
10=
6=
2
1
3
6
4
3
1
4
5
20=
10=
1
1
9
2
2
7
3
8
3
3
6
4
5
7
4
5
6
8
4
7
5
6
Patterns for the number of triangles in a polynomial of n+1 sides:
In general, we count:
Triangles formed
with the vertices
1, 2, ..., n+1
Triangles formed
with the vertices
1, 2, ..., n
Triangles formed
with the vertex
n+1 and a pair of
other vertices among
1, 2, ..., n
More generally:
Suppose there are n+1 people in your class. In how
many ways can a group of k+1 people be chosen?
Ways to choose
a random group
of k+1 people
Ways to choose
a group of k+1
people if you’re
not included
Ways to choose
a group of k+1 people
if you’re not included
(it remains to choose k
other people from the
remainder of the class)
Think about: Sums