Pascal's Triangle

Pascal’s Triangle is a famous triangular array of numbers with several interesting properties.  Each line of the triangle starts and ends with the number 1, and the numbers in between can be obtained by finding the sum of the two numbers above it.  The triangle can then be continued indefinitely.

                            

						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

Pascal’s Triangle

Unfortunately, Pascal’s triangle is rarely mentioned in high school math curriculum, except as a footnote for expanding binomials, because the coefficients for the expanded form of (x + y)ⁿ match up with the rows of Pascal’s Triangle.  For example, (x + y)² = x² + 2xy + y² (1, 2, 1), (x + y)³ = x³ + 3x²y + 3xy² + y³ (1, 3, 3, 1), and (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴ (1, 4, 6, 4, 1).  But there are many other unique properties to the triangle.

First of all, the rows of Pascal’s triangles add up to a power of 2.  For example, the numbers in the third row (1, 2, 1) add up to 4 which is 2², the numbers in the fourth row (1, 3, 3, 1) add up to 8 which is 2³, and the numbers in the fifth row (1, 4, 6, 4, 1) add up to 16 which is 2⁴.

						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

The rows of Pascal’s Triangle add up to powers of 2.

Second, the diagonals of Pascal’s Triangle represent totals for different simplexes (lines, triangles, tetrahedrons, etc.).  For example, the third diagonal row (1, 3, 6, 10, 15, etc.) are triangular numbers, because objects in these amounts can form equilateral triangles, and the fourth diagonal row (1, 4, 10, 20, etc.) are tetrahedral numbers, because of objects in these amounts can form tetrahedrons (triangular pyramids).

						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

The diagonals of Pascal’s Triangle represent totals for different simplexes.

Triangular Numbers: 1, 3, 6, 10, etc.

Tetrahedral Numbers: 1, 4, 10, 20, etc.

Third, another diagonal of a different slope in the numbers of Pascal’s triangle add up to different numbers in the Fibonacci sequence.  As you might recall, the numbers in the Fibonacci sequence can be obtained by adding the two numbers before it, so starting with two ones, it is 1, 1, 2, 3, 5, 8, 13, etc.  Numbers in the pictured diagonals (and parallel diagonals) of Pascal’s triangle add up to the numbers in the Fibonacci sequence.  For example, 1 = 1, 1 = 1, 1 + 1 = 2, 1 + 2 = 3, 1 + 3 + 1 = 5, 1 + 4 + 3 = 8, 1 + 5 + 6 + 1 = 13, etc.

						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

These Pascal Triangle diagonals add up to different numbers in the Fibonacci sequence.

Fourth, the numbers in the Pascal’s triangle represent the number of combinations that are possible in a set, which is useful in statistics.  For example, if you wanted to know how many different possible two-card hands you can be dealt from a deck of five cards, it would be ₅C₂ = 10, which is 5 + 1 = 6^th row, and 2 + 1 = 3^rd number in Pascal’s Triangle.  (If the five cards are A, B, C, D, and E, then all the possible two-card hands are AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE, which makes ten total possibilities.)  The number of different possible four-card hands you can be dealt from a deck of six cards would be ₆C₄ = 15, which is the 6 + 1 = 7^th row, and 4 + 1 = 5^th number in Pascal’s Triangle.

0C0

1C0

1C1

2C0

2C1

2C2

3C0

3C1

3C2

3C3

4C0

4C1

4C2

4C3

4C4

5C0

5C1

5C2

5C3

5C4

5C5

6C0

6C1

6C2

6C3

6C4

6C5

6C6

Combination Sets in Pascal’s Triangle

Finally, as mentioned in an earlier blog post, the number of endpoints, line segments, faces, and solids of each multi-dimensional simplex corresponds to a row in Pascal’s Triangle.  For example, the 1-D simplex (line segment) is comprised of 2 endpoints and 1 line segment (2, 1); the 2-D simplex (triangle) is comprised of 3 endpoints, 3 line segments, and 1 face (3, 3, 1); the 3-D simplex (tetrahedron) is comprised of 4 endpoints, 6 line segments, 4 faces, and 1 solid (4, 6, 4, 1), etc.

3D Simplex: Tetrahedron

Comprised of 4 endpoints, 6 line segments, 4 faces, and 1 solid

Pascal’s Triangle has some amazing unique properties.  It can be applied to the expansion of binomials, powers of two, sequences of numbers, statistics, and properties of simplexes.  It truly is a wonder in mathematics!

Vectors for a Given Angle and Vector

As a high school math teacher, I often have to come up with my own problems to give on worksheets and tests.  Many times I will take an existing problem from the textbook and change the numbers, but the trick is to change the numbers in such a way so that the level of difficulty of the problem is maintained.  To accomplish this, I often have to work the problems backwards from the answer to get to the question.  For example, a typical factoring problem would be to factor x² + 5x + 4.  To solve this, you find two numbers that multiply to 4 and add up to 5, which are 1 and 4, and so the solution is (x + 1)(x + 4).  If I wanted a problem similar to this I can’t just randomly change the numbers in x² + 5x + 4, because the solution might have a fraction or radical or both, which would make the problem more difficult.  However, I can randomly change a number in the solution and then work backwards to get a new problem of the same difficulty.  So if I wanted the solution to be (x + 1)(x + 5) instead, I would work backwards by multiplying it out and get x² + 6x + 5 as the new problem.

Working backwards may be relatively simple for factoring, but for other problems, such as finding the angle between two vectors, it can be more difficult.  A typical problem in trigonometry would be to find the angle between the vectors (4, 3) and (3, -4).  Using tan q = ^{m1 – m2}/_{1 + m1m2}, with m₁ = ³/₄ and m₂ = ^-4/₃, q = tan^-1(^25/12/₀) = 90°.  

But what if you want the solution to be a different angle, like 45°?  What other vector makes a 45° angle with (4, 3)?  Asking the question backwards like this does not make for a typical trigonometry problem.

To find vectors for a given angle and vector, we need rearrange the equation tan q = ^{m1 – m2}/_{1 + m1m2}.  Multiplying both sides by 1 + m₁m₂, we get tan q (1 + m₁m₂) = m₁ – m₂.  Distributing, we get tan q + tan q m₁m₂ = m₁ – m₂.  Rearranging terms, we get m₂ + tan q = m₁ – tan q m₁m₂.  Factoring out m₁ we get m₂ + tan q = m₁ (1 – tan q m₂), which means m₁ = ^{m2 + tan q}/_{1 – tanq m2}.  A similar argument rearranges tan q = ^{m1 – m2}/_{1 + m1m2} to m₂ = ^{m1 – tan q}/_{1 + tanq m1}.  Since the vectors can be interchanged without changing the angle between them, we can combine both formulas to make m₁ = ^{m2 ± tan q}/_{1 ∓ tan q m2}.  Therefore, to find a missing vector of a given angle and another vector, we can use the formula m₁ = ^{m2 ± tan q}/_{1 ∓ tan q m2}, where q is the angle and m₁ and m₂ are the slopes of the two vectors.  This means that the other vector that makes a 45° angle with (4, 3) will have a slope of ^{m2 ± tan q}/_{1 ∓ tan q m2} = ^{3/4 ± tan 45°}/_{1 ∓ tan 45° 3/4} = 7 or ^-1/₇.  A few vectors with these slopes can include either (1, 7) or (7, -1).