Patterns and Combinatorics

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Presentation 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 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? 2 1 1 2 1 1 2 5 3 6 3 4 2 3 3 4 4 5 1 1 2 1 2 9 2 7 3 8 3 8 3 6 4 7 4 7 5 6 4 5 5 6

We counted all lines connecting the vertices of the polygons: Can you explain the pattern? 1 2 1 1 2 1 5 2 3 6 4 3 3 2 3 4 4 5 3 lines 6 lines 10 lines 15 lines +3 +4 +5 2 1 1 1 9 2 2 7 3 8 3 8 3 4 6 7 4 7 5 5 6 6 4 5 +6 +7 +8 21 lines 28 lines 36 lines

The number of lines connecting the vertices of a polygon with n sides is the (n-1)th triangular number: 2 1 The n-th vertex contributes n-1 red lines: 3 4 counts how many pairs of 2 vertices we can form from n given vertices. Each pair determines a line. also denoted

Count the rows in Pascal’s triangle starting from 0 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= 2nd slanted row 1= 1= 2= 1= 3= 3= 1= 1= 4= 6= 4= 1= 5th horizontal row 1= 5= 5= 1= 10= 10=

Any three vertices of a polygon form a triangle Any three vertices of a polygon form a triangle. How many such triangles are there in each polynomial? 2 1 1 2 1 1 2 5 3 6 3 4 2 3 3 4 4 5 1 1 2 1 2 9 2 7 3 8 3 8 3 6 4 7 4 7 5 6 4 5 5 6

Triangles in each polynomial: 1 2 1 1 2 1 5 2 3 6 4 3 3 2 3 4 4 5 1= 4= 10= 20= 3= 6= 10= 2 1 1 1 9 2 2 7 3 8 3 8 3 4 6 7 4 7 5 5 6 6 4 5

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