Going through some old class notes from a few years ago and rememered this old post from 2008....
Hope all these kids are doing well today.
In my B2 Pre-calc class today, we re-discovered a theorem about
Pascal's triangle that I had not known. It began, appropriately enough,
with a question Jacob C. asked about dealing cards from a standard
deck; "How many 13 card hands can be dealt that contain exactly two
suits. " As we were working through the problem, I began by attacking
the somewhat easier problem, "how many hands can be dealt with only
hearts and diamonds, but at least one of each." We began writing out
the possibilities of 12 hearts, 1 diamond plus 11 hearts two diamonds
..etc To make life easy, for this short while let’s let (n,r) mean “n
choose r” , the combinations of n things taken r at a time. So we
needed to find (13,1)(13,12) for the first part, 12 hearts and 1
diamond. Then we needed to add on (13,2)(13,11) + (13,3)(13,10)…. And
all the way down to (13,1)(13,12). One of the clever ones quickly
realized that each of these pairs were just the same number due to the
symmetry of Pascal's triangle, and so we were really looking for (13,1)2 + (13,2)2...
etc. While some of the kids were adding these on their calculators, I
wrote out several lines of the arithmetic triangle and began to write
the sums of squares on the right....
I wrote the totals of each row, 1, 2, 6, 20, 70.. it struck me that
they were all the center number of an even numbered row, (2n,n). I
remembered them from working with Catalan’s Numbers (another cool
pattern that shows up in Pascal’s triangle). About the time the first
students were coming up with an answer, I asked them to check (26, 13)
and compare it to the answer they got for the actual squares of the
Close, but not right, was the reply....
huh??? … , oh yeah, we had avoided the case of (13,0) and (13,13)
because we wanted to ignore the case where all were hears or all were
diamonds, so the answer to our mini-problem was (26,13) - 2; and the
only thing needed to solve the original problem was to multiply by 6,
to account for all the ways we could pick two suits to be in the hand
out of the four possible suits.
When I showed them
the result, and we checked a couple of more cases to be more sure, I
admitted that I had never seen this theorem. One kid suggests it should
be a test question… I countered with, “and extra credit for the person
who comes up with the best name for it. Several played to my ego,
“Ballew’s theorem, of course!” but then they thought they might deserve
partial credit, and hence the name, B-2 theorem, at the top.
we were not the first to stumble across this little gem. I haven’t had
time to chase it down fully, but it may actually date back to the
Chinese around the 12th century. So fame and fortune will have to wait,
but when you walk in the footsteps of greatness, you’re taking pretty
big steps; so congratulations class, I’m proud of you, and it will
always be the B-2 theorem when I teach it. Dennis was going to send me a
class picture we took on his phone, so if it turns out, I will add that
While I was searching for the history of the
sum of the binomial coefficients, I came across another place where the
triangle is related to squares. One of those theorems we teach when we
get to sequence and series in high school is the sum of the integers, 1 +
2 + 3 + … + n, and the sum of the squares of the integers, 12 + 22 + 32 + ….. + n2. Usually we present the formula
for this last without proof since it occurs before they are introduced
to inductive proofs. As I was researching I came across this neat
little relation to the arithmetic triangles. To find the sum of the
squares of the first ten integers, just go down to 10 at (10,1) and turn
right and follow the diagonal down two numbers to (12,3) and add this
to the number on the diagonal above it (11,3), the sum of 220 + 165 =
385 which is the same as 12 + 22 + … + 102
In general you can find
And I think they can accept that as evidence, at least until we get to inductive proofs.