Counting School Supplies for the First Days of School

Date: 07/23/97 at 15:53:19
From: Anonymous
Subject: Triangle number question

Dear Dr. Math,

I am a sixth grade middle school teacher teaching math as a lone 
subject for the first time. I found a terrific activity for the
start of the school year, but I don't understand the "rule" entirely. 

The problem is counting school supplies for the first twelve days of
school - like the 12 days of Christmas.

I am not sure of triangle numbers but the rule is n(n+1)/2 for each 
day and n(n+1)(n+2)/6 for the entire sequence.  I know it works 
because I did the math. My question is, where did the divide 
by 2 and 6 come from? I know one of my students will ask why you 
divide by 2 and 6 and I can't figure it out.
 
Any help you could give me in understanding this would be greatly 
appreciated.  

Thank you in advance for your help.

Wendy Bender

Date: 07/24/97 at 13:53:01
From: Doctor Rob
Subject: Re: Triangle number question

We are pleased that you want to give correct reasons to your students.
This is admirable.

Consider this picture (I'll only draw it for n = 5, but you can do 
this for any size):

   ___ ___ ___ ___ ___ ___
6 |   |   |   |   |   | X |
  |___|___|___|___|___|___|
5 |   |   |   |   | X |   |
  |___|___|___|___|___|___|
4 |   |   |   | X |   |   |
  |___|___|___|___|___|___|
3 |   |   | X |   |   |   |
  |___|___|___|___|___|___|
2 |   | X |   |   | Y |   |
  |___|___|___|___|___|___|
1 | X |   |   |   |   |   |
  |___|___|___|___|___|___|
    1   2   3   4   5   6


Every box in this diagram is specified by a pair of integers (a, b), 
where each is at least 1 and at most 6.  The ones marked with X are 
represented by the pairs (a, a).  For example, the box marked with Y 
is (5, 2), since it is in the fifth column and second row.

The boxes with a > b are the ones below the diagonal marked with Xs.  
They are arranged in a triangle, and their number is the 5th 
triangular number, 5*(5+1)/2 = 15. There are (n+1)^2 boxes in all.  
We throw away the n+1 boxes marked with Xs, leaving n^2 + n = n*(n+1).  
They correspond to all pairs (a, b) with a unequal b. Half of these 
are above the diagonal (have a < b) and half are below the diagonal 
(a > b).  

To count only one of the two triangles, we divided by 2 = 2*1. This 
corresponds to the two different pairs you can make out of two unequal 
numbers, (a, b) and (b, a), of which you are picking only one. The 
result is n*(n+1)/2, as you know.

For pyramidal numbers (sums of triangular numbers), a similar thing
happens, but now we have a three-dimensional cube of boxes, each 
specified by a triple (a, b, c), telling you in which column, row, and 
plane it lies. Since there are six ways to rearrange three unequal 
numbers, and you only want to look at the one of those with a > b > c 
which lie in the pyramid, you have to divide by 6 = 3*2*1. 

You start with (n+2)^3 numbers and throw out all boxes with a = b or 
a = c or b = c or a = b = c. There are (n+2)^2 triples with a = b, and 
the same number with a = c, and likewise for b = c, so we subtract 
3*(n+2)^2 from (n+2)^3. We have subtracted too much, however, since 
we have deducted the (n+2) triples with a = b = c three times. As a 
result, we have have to add 2*(n+2) back in to get the correct answer. 
This gives (n+2)^3 - 3*(n+2)^2 + 2*(n+2), which, when simplified, 
yields n*(n+1)*(n+2). That is the number of triples of unequal numbers 
from 1 to n+2.  We have to divide that by 6 to get the number of 
triples with a > b > c, as noted above.

There are 5 other copies of the pyramid in the cube, too.  They 
satisfy a > c > b, b > a > c, b > c > a, c > a > b, and c > b > a,
respectively.  This may be hard to draw on the blackboard!  
Nevertheless, they are there.

Good luck!  I hope this helps.  If not, write back and we'll try again 
to explain.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/