Consider a triangular pyramid of balls, constructed by placing each ball (except those on the bottom) in the hollow formed by the three balls under it. If such a pyramid has 12 balls on the side of the bottom level, how many balls in total does it contain?

I recently recieved the following message in my e-mail:

> Dear TomM,
>
> I was interrested in your solution to the problem of
> the pyramid of balls. As it happens, I was recently
> considering this issue and came up with the same
> soution. However I suspect that there is possibly
> another aspect to this worthy of consideration. If
> we look at triangular numbers alone, i.e. the
> solution to how many balls "y" would make up a
> triangle of "x" sides then the equation is :
>
>
> y = x * (x + 1)/2
>
>
> If we extend this to the third dimension i.e. how
> many balls make up a triangular pyramid of "x" sides
> then the sulution is :
>
> y = x * (x + 1)/2 * (x + 2)/3
>
> (or as you note : x * (x + 1) * (x + 2)/6)
>
> One could be pedantic and suggest that the first
> term in the equation (i.e. the intial x) is actually
> (x + 0)/1giving a formula for the pyramid of :
>
> y = (x + 0)/1 * (x + 1)/2 * (x + 2)/3
>
> It is interresting therefore to speculate on a
> general formula for the number of balls contained in
> an "n" dimensional "triangular-type object"
> (whatever that would look like) with the general
> formula :
>
> y = (x + 0)/1 * (x + 1)/2 * (x + 2)/3 * ....... (x +
> n - 1)/n
>
> Regards,
>
> Jim G.

Here is my reply:

An interesting idea. It's hard to visualze four or more dimensions, so it's had to be sure, but it looks like the relationship holds.

If we call the function which gives the value y for n dimensions Tn(x),
then T1(x) = x = (x+0)/1!
T2(x) = x(x+1)/2 = (x+0)(x+1)/2!
T3(x) = x(x+1)(x+2)/6 = (x+0)(x+1)(x+2)/3!
...
Tn(x) = x(x+1)(x+2)...(x+n-1)/n!

If we let X = x+n-1, the numerator becomes (X)(X-1)(X-2)...(X-n+1) = X!/(X-n)!

So Tn(x) = X!/(X-n)!(n!) = C(X,n) where C is the "combinatorial" operator more usually designated by placing the two arguments vertically between a set of parenthases(very hard to accurately type in an e-mail*)

So Tn(x) = C((x+n-1),n)

*or in a message box

Posted by TomM on 2003-07-04 14:41:56