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I have a flat cone in which I can drop marbles. There is space for one marble in the bottom row, two marbles in the row above that, three marbles above that, and so on.

\         /
 \5      /
  \3 6  /
   \2 4/
    \1/

A marble with the number $x$ means it was the $x$th marble to be dropped. The marbles follow the laws of gravity, which means that a given position can be occupied by an incoming marble only if another marble (or a wall of the cone) exists to both its bottom left and bottom right.

The marbles can pile over the top of the cone like this:

      8
\    7 4/
 \  6 3/
  \5 2/
   \1/

but, in this case, the maximum number of marbles you can put is $16 = 4 \times 4$. In general, you are allowed to put $ab$ marbles where $a$ and $b$ are the lengths of the sides of the cone.

My question is, if the sides of the cone are of length $a$ and $b$, then in how many ways can you drop all the $ab$ marbles into the cone? This could possibly be solved more specifically for $a = b$, or more generally for three-dimensional cones.

(This question is inspired by this one, which is (in my opinion) the special case $a = b = 3$.)

Edit: Since the flat cone case is in easy bijection to standard Young tableaux as pointed out by @Marcel in the answer below, can anybody suggest some approaches for three-dimensional $k$-gonal pyramids? Just tetrahedrons, maybe?

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shardulc
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1 Answers1

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Your marble arrangements are in easy bijection to the famous standard Young tableaux, with restrictions.

Young tableaux are defined for integer partitions $\lambda=(\lambda_1,\lambda_2,...)$. The number of parts is denoted $\ell(\lambda)$. The sides of your cone are the first part, $a=\lambda_1$ and the number of parts $b=\ell$. Since you must put $ab$ marbles, the partition in your case has $b$ parts, all equal to $a$. This is usually written as $\lambda=(a^b)$.

The number of standard Young tableaux for given $\lambda$ is denoted $f^\lambda$. You can see an explicit formula here: https://en.wikipedia.org/wiki/Hook_length_formula

The quantity you are looking for is given by $$f^{(a^b)}=(ab)!\prod_{i=1}^b\frac{(b-i)!}{(a+b-i)!}.$$

Marcel
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  • Thanks for introducing me to Young tableaux. Wouldn't the quantity I'm looking for be the Hook length formula applied to the partition $(a,a,a,\dotsc,a)$ (with $b$ terms)? Also, what are those $\delta$ symbols, and why are the terms being summed in your expression? – shardulc Mar 12 '16 at 04:21
  • I'm also trying to generalize to a three-dimensional tetrahedron, square pyramid, or $k$-gonal pyramid where the marbles follow the same rules, but the number of marbles in the $n$th layer is the $n$th $k$-gonal number. Do you know of any existing literature on this topic? – shardulc Mar 12 '16 at 04:27
  • I have edited the answer (the final bit was correct but unnecessarily complicated; never mind). The only 3D analogue I know are related to so-called "plane partitions". I don't know much about them, but you can look them up. – Marcel Mar 12 '16 at 13:21