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How many Young diagrams are there with length of at most $p$ rows and $q$ columns, without restrictions on the weight of the diagrams?

The previous 2 questions asked for

  1. Total number of Young diagrams with weight 6 and

  2. Total number of Young diagrams with weight 7, excluding diagrams with more than 3 rows.

The answers I got are 1) 11 and 2) 8,

By counting the number of partitions of the integers 6 and 7 respectively, and subtracting $7, 6+1, 5+2, 5+1+1, 4+3, 4+2+1, 4+1+1+1$ from all partitions of 7.

However, I have no idea how to do that with unspecified weight, and taking into account the question details.

Notice the question is talking about Young diagrams, as opposed to Young tableaux.

Jyrki Lahtonen
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John Doe
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  • I think in effect you want the number of solutions to $0 \le x_1\le x_2\le \cdots \le x_p \le q$. It might be worth adding something to each to make them distinct, at which point this becomes easy – Henry Jan 13 '23 at 22:27
  • @Henry how do I set everything up? – John Doe Jan 15 '23 at 13:33
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    An even easier approach would say you want a path from one corner of the $p\times q$ rectangle to the opposite corner to separate your Young diagram from the rest of the rectangle – Henry Jan 15 '23 at 16:41

1 Answers1

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Wanted to post a complete answer in case someone finds this question later. This follows up on Henry's comment of 15 January.

First, judging from your answer to #2, it looks like you're visualizing partitions in the "French" convention, with $4+2+1$ consisting of three columns with heights 4, 2, 1 from left to right. (The other option is "English" where $4+2+1$ corresponds to three rows with lengths 4, 2, 1 from top to bottom.) The enumerations don't depend on this choice, of course, but I wanted to answer your question in the appropriate context.

A partition with at most $p$ rows and at most $q$ columns is determined by a path starting at $(0,p)$ and ending at $(q,0)$ consisting of $p$ steps down and $q$ steps right. One extreme case is $q$ rights followed by $p$ downs, giving the partition $p+\cdots+p$ repeated $q$ times of weight $pq$. The other extreme case is $p$ downs followed by $q$ rights, the empty partition of 0. Supposing $p$ and $q$ are large enough, the path given by $p-4$ downs, right, down, down, right, down, right, down, $q-3$ rights describes $4+2+1$.

The number of such paths is the binomial coefficient $\binom{p+q}{p}$ since it is enough to specify the positions of the $p$ down steps among the $p+q$ total steps. (Generally, questions about a specific weight are harder.)

Brian Hopkins
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