Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

  • Advertisements

Archive for the ‘Pascal’ Category

That Rascal Pascal

Posted by mark schwartz on May 30, 2016


This piece came about because one particular Contemporary Math class became intrigued with Pascal’s Triangle and collectively asked if there wasn’t more too it. I’m not sure what they meant by “more to it” but when we talked about it, they said that patterns were interesting. I said I knew a few things and would look into a little more. In the next class, one student said that she had “googled” Pascal and found a heap of stuff. She presented some to the class and we played with the patterns some more. But none of her findings engaged them like the story below, particularly the endnote.

The Story:

Of the many reasons that Blaise Pascal is remembered, the principle one is “Pascal’s Triangle”. Of course, it’s not really his. Several Chinese mathematicians had discovered and explored it long before Pascal. He popularized it. Legend also has it that he invented the roulette wheel, which is consistent with why Pascal’s Triangle exists. And if you want to know more – for example Pascal’s Wager – consult the web. But for the Triangle …

It describes a basic pattern that emerges when expanding a binomial to any power. For Example (a + b)2 gives a2 + 2ab + b2, while (a + b)4 gives a4 + 4a3b + 6a2b2 + 4ab3 + b4.

The pattern of interest is the coefficients of the terms. The essence of the pattern can be seen in the pattern in which the powers are aligned with the coefficients. The power, n, is the exponent of the binomial (a + b)n. The Triangle through the 5th power is:


0                              1

1                            1      1

2                        1      2      1

3                    1     3       3     1

4                 1    4      6       4    1

5              1    5    10    10     5    1

Anyone familiar with this pattern can generate the next line, which would be (a + b)6. Then again, even if you’re not familiar with the pattern, after several moments of looking at this “Pascal’s Triangle” you might see the pattern, which is: the first and last terms are 1; and any number in any row is generated by adding the two numbers above it and slightly to the right and left. For example, for the fourth power, the middle number 6 comes from adding the 3 and the 3 from the row above; the second 10 in the fifth power comes from adding the 6 and 4 from the row above, etc.

This is one of the interesting patterns in Pascal’s Triangle. This is how to generate the next power from the previous one. There is also a way of determining a particular term in an expansion. In expanding (a + b)4  there is a method for utilizing the values from the previous term to generate the coefficient and exponents of the following term. In essence, any “row” in this triangle are the coefficients of the terms in the expansion based on the power.

In the example above, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 and looking only at the coefficients you can see the pattern. A little demonstration will show the beginnings of one of the features that is rarely noted.

The coefficient of the first term is 1 and the exponent of the first term is 4. Multiply the coefficient by the exponent of “a” and dividing this by the number of the term gives the coefficient of the second term, 4. The coefficient of the second term is 4 and the exponent of “a” is 3. Multiply the coefficient by the exponent and dividing this by the number of the term (2) gives the coefficient of the third term, 6. This works for the rest of this row and it works for any row. It’s a powerful pattern which is a lot simpler than actually multiplying out the terms in a binomial expansion.

But notice what really happens using this pattern. The coefficient of the second term, before actually doing multiplication or division is  4/1 (the multiplication of 1×4 in the numerator is considered trivial, so the “1” is not noted). The coefficient of the third term, including the second term, is . Continuing this for the 4th and the last term, the coefficients for the terms in (a + b)4 are:

1a4 + 4/1a3b + (4•3)/(1•2)a2b2 + (4•3•2)/(1•2•3)ab3 + (4•3•2•1)/(1•2•3•4)b4

If you’re familiar with factorials, you will recognize that the coefficients for the 2nd through 5th term are factorials. But, together the numerator and denominator are really how combinations are generated. The symbol nCr  indicates how to find  how many combinations there are of “n” things taken “r” at a time, and the formula is  n!/(r!(n  ̶  r)!) … (a factorial n! means n(n-1)(n-2)…1).

The coefficient of the third term looks like the formula and actually is the formula; you just have to realize that “n” is 4, “r” is 2 and “(n−r)” is 2. The (n-r)! term in the denominator cancels the 2•1 part of 4! in the numerator.

So, in essence, when finding the coefficients of the terms in a binomial expansion, or when generating the Pascal’s Triangle, what is really being determined is a series of combinations.

But, here’s the fun! Using the coefficient of the fourth term, which is 4, what it describes really are how many combinations there are of ab3, the variable part of the term. Recall that with combinations, any like terms are only counted once. For example, “ab” is the same as “ba” if combinations are being counted, but they are not the same if permutations are being counted (order counts in permutations).

So look at the combinations of ab3, which can also be written as “abbb”. The four combinations are “abbb”, “babb”, “bbab”, and “bbba”.  The coefficient of any term is the numerical value of the number of combinations. An example: the third term of the expansion of (a + b)4 is 6a2b2, noting that there are 6 combinations of 4 terms taken 2 at a time and they are aabb, abab, abba, baab, baba, and bbaa.

This, of course, makes sense. But what of a term like 7a3b2? Is the “7” correct and if not, what should the number be? You can create the entire expansion of (a + b)5 until you get to the  a3b2 term, or given the relationship provided by Pascal you can realize that the combination by using the combination formula is 5!/(3!2!) which gives 10. You can go one step further into the abyss, and simply provide a term without a coefficient (or a coefficient of 1), and ask for the coefficient. For example, if the term is x3y3, the coefficient would be … ?     (the answer is 20). By the way, you can have a term like 7a3b2 but we’re talking Pascal here, not general Algebra.

There’s another fun event that Pascal provides. What follows was a summary handout to the students because I had them do this activity another way (see the Endnote below). This same finding-the-coefficient activity applies to answering questions about the following diagram:

M   N    O    P

I     J     K     L

E    F    G     H

A    B    C    D

The question: being allowed to move only north and only east in any order from letter to letter, how many paths are there to get from A to P?  Tracing all the possible paths is somewhat nightmarish and takes a lot of careful counting and recording, but Pascal to the rescue!!

Counting north from A to the top of the square gives 3, and counting east from A to the extreme right is also 3. Using these values in the combination formula gives  6!/(3!3!)  or 20. Tracing and keeping track of this many paths wouldn’t be easy; better to use the formula.  Try finding how many ways there are to get from E to O.  It’s “up ? and over ?”, so the formula would look like  ?/(?!?!) and this equals ?.

The answer is 6, and in this case you could verify this by actually tracing the paths … or use nCr.


When I presented this path tracing problem in class, I had them go out to the hallway, which had checkerboard tiling and had them move from tile to tile, having designated that they should mark off a 4by4 square, and of course, some groups decided to do a 5by5. After “walking” through some easy paths, I asked them to do the extreme case (A to P), which they started but then realized that counting and recording the paths was – as one student said – “a crazy way to do it” – to which I replied “how else”? Someone eventually had an “aha” moment and shared it with everyone. The next time we had class, several students said that they can’t walk the halls without thinking about the path from point to point … an unexpected and interesting outcome for them and me.


Posted in algebra, binomial expansion, combination/permutation, math instruction, mathematics, Pascal | Leave a Comment »

A Toddler, Pascal and Fibonacci Climb Steps

Posted by mark schwartz on April 14, 2016


I once taught a Contemporary Math class, designed as a survey course for those whose major required one math course but nothing heavy like a full Algebra, although some students had Pre-Algebra or Algebra in high school. Among the topics were binomial expansion, Pascal and Fibonacci. A student asked if these guys knew each other and when I asked him what he meant he said “I thought most of these guys stuff was connected somehow”. It wasn’t an elegant question but it turned out to be insightful. One of the exercises I had the class do was “a toddler climbs the steps”. As I thought about his question and the exercise and played with it some more I came to see a connection between these guys stuff. As a class we discussed our way through the story that follows and learned considerably more because his question led us to it.

The Story

Let’s start with a set of steps and a toddler. The toddler is about to climb a set of steps but is limited to two ways; either 1 step at a time or 2 steps at a time, or a combination of 1-step and 2-step climbing.

The question is how many different ways the toddler might climb a fixed number of steps. What this means, for example, is … suppose there are only 2 steps. The toddler could climb 1 step at a time, which can be represented as 1.1, for a total of 2. The toddler could also simply climb the 2 steps once.

Continuing this, if the toddler were to climb 3 or 4 or 5 etc., the number of ways to climb can be determined by simply mapping the possibilities and counting them. To clarify the entries in the “ways-to-climb” column in the table below, take a look at the ways-to-climb for 3 steps. “1.1.1” means that the baby climbed them 1 at a time. “1.2” means that the baby climbed one step followed by a 2-step climb and “2.1” means the baby first climbed 2 at a time, followed by climbing one. These are the only 3 ways that the baby could climb 3 steps.

# steps           Ways to climb                                                                                                       Total

1                        1                                                                                                                        1

2                        1.1 2                                                                                                                   2

3                        1.1.1     1. 2     2.1                                                                                              3

4                  1.1.2   1.2.1   2.1.1   2.2                                                                         5

5                 2.2.1    2.1.2    1.2.2                         8

6                 2.2.2                                                                      13

An interesting and familiar pattern shows up in the “Total” column. It’s the Fibonacci series. But the question is “why’? This question can be answered by looking first at the relationship between Pascal’s triangle and the Fibonacci series.

Pascal’s triangle is a series of rows showing the coefficients of the terms that result from the expansion of the binomial

(x + y)n using n = 0 through 6. which will generate 7 rows. Pascal’s triangle, left aligned, is:


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 Fibonacci series is found in the triangle by starting at a “1” in the left-hand column, adding the numbers as you move diagonally up left to right at about 45 degrees. For example;

In the fifth row, moving diagonally up, there is a 1, 3, 1, totaling 5.

In the sixth row, moving diagonally up, there is a 1, 4, 3, totaling 8.

In the seventh row, moving diagonally up, there is a 1, 5, 6, 1, totaling 13.

These are a sequence in the Fibonacci series.

But the pattern goes even deeper. Pascal’s triangle only shows the coefficients of the terms in the expansion. For example, the expansion of ( x + y )3 is   x3 + 3x2y + 3xy2 + y3, which are the coefficients in the fourth row of Pascal’s triangle. What the baby climbing the steps does is show the pattern, somewhat hidden, of the expansion and further, shows how the Fibonacci series can be seen in the triangle.

Let’s begin with the number of steps = 2. In this case, and in all subsequent cases, let 1 step = x and 2 steps = y. For 2 steps then the outcomes in the ways-to-climb column would be x.x and y, not 1.1 and 2. Here we will use exponential notation for x.x giving x2 and this will be done for all subsequent cases.

So, for 2 steps that has 2 outcomes, they are x2 and y. Now look at Pascal’s triangle and the Fibonacci series that can be generated from it.

An expansion of ( x + y )1 gives           x + y

An expansion of ( x + y )2   gives        x2   +   2xy +   y2

An expansion of ( x + y )3 gives          x3 +  3x2y +  3xy2 +  y3

An expansion of ( x + y )4 gives          x4 +   4x3y + 6x2y2 + 4xy3 +  y4

The expansions have been “spaced” so that you can see that if you add the coefficients of x2 in the second expansion and the coefficient of y in the first expansion you get 2. If you add the coefficients x3 and 2xy you get 3. If you add the coefficients of x4, 3x2y and y2 you get 5 and 2, 3, 5 are a sequence in the Fibonacci series.

Below in rows 1 and 2 is an example showing the generation of the relationship between “ways to climb” when the toddler climbing 5 steps. Row two is the first term of (x + y)5 , the second term of (x + y)4 and the third term of (x + y)3. This is the same as what you get moving diagonally up in Pascal’s triangle.

Row 1. Ways to climb                2.2.1   2.1.2    1.2.2

Row 2. xy form                   x5                                         4x3y                                                     3xy2

If you add the coefficients of the terms in row 2 above you get 8, which is the same as finding the Fibonacci series from Pascal’s triangle beginning in the sixth row of the above Pascal’s triangle. What’s more, row 2 demonstrates the relationship between the coefficients in Pascal’s triangle, the number of ways the toddler can climb steps, the terms in a binomial expansion and finally, the Fibonacci series. You can see that the entries in row 1- each term having 1s and 2s – are, in essence, the exponents of x and y in row 2, where 1 step = x and 2 steps = y. At the beginning of this piece, I noted that you could find the number of ways by mapping it, but given climbing steps is connected to the binomial expansion, use the binomial expansion. For example, if there were 10 steps, in how many ways could the toddler climb these 10 steps?

Here’s how get the number of steps; generate a series of combinations using nCr because that’s how you can get the coefficients of the terms in a binomial expansion. For example given x2y2, 2C2 gives 6. But it’s not a single calculation.

Knowing that the Fibonacci series is the “Total” outcome as seen in Table 1 and also that the Fibonacci series is the outcome by moving diagonally up at 450 in Pascal’s triangle, the nCr calculation is a series of calculations using the following pattern:

Start with 10C0 and generate the series by decreasing “n” by 1 and increasing “r” by 1 each time. In this case, the next two terms would be 9C1 followed by 8C2, etc..

The question then is “how do you know when to stop”? At some point in generating this series, r will be greater than n and since this is illegal (your calculator will tell you!), you’re done. Just add up the values that have been generated. Given 10 steps, there are 89 different ways for the toddler to climb them. This may seem awkward, but as I indicated earlier, you could map the outcomes but in this case and with larger numbers it would be very time consuming and very trying to map it. Stick with nCr.

As I noted in the introduction we learned a lot as we worked through this. More importantly though, in my opinion, was the engagement of the class in doing the work and seeing all the connections. It didn’t exactly follow the curriculum, but it did follow a student’s imaginative question.

You can leave a comment below if you’d like to … it would be appreciated.










Posted in Category One, Category Three, Category Two, Fibonacci, Pascal | Tagged: | Leave a Comment »