Marveling At The Historical

Math Oldies But Goodies

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    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.

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Archive for the ‘combination/permutation’ 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.


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