Number of ways to complete a partial Young tableau

Question

Suppose we have a Young tableau with missing entries. Then there can be many number of ways we can complete the Young Tableau.

Is there any specific method to find the number of ways we can complete a partial Young tableau?

For example if we have the following partial Young tableau, how many ways we can complete it?

Need some hints. Thanks.

Answer 1

I don't believe that there's a nice general way to do such problems.

Your example is fairly easy, though. You put the 1 in the only place it can go for a standard Young tableau. To count how many standard Young tableaux have 9 at the end of the second row, use the hook length formula (mentioned in Jean Marie's comment) on the remaining shape $(4,2,1,1)$ which gives $8!/(7\cdot4^2\cdot2^2) = 90$ tableaux.

To test this idea, there are 3 possibilities for where 9 can go: in addition to the end of the second row that you gave, it could be at the end of the first row or at the bottom of the first column. Using the hook length formula on $(3,3,2,1)$ and $(4,3,1)$ gives 56 and 70 possibilities, respectively. The sum $90+56+70 = 216$ corroborates with the total count for standard Young tableaux of shape $(4,3,1,1)$. Let me show these counts as

$$\begin{matrix} 0 & 0 & 0 & 56 \\ 0 & 0 & 90 \\ 0 \\ 70 \end{matrix}$$

Things are more challenging for, say, counting the possibilities of 5 in a specified position. Consider putting 5 in the first row, second column. That gives just one way to place 1, 2, 3, and 4: in the first column, in order.

$$\begin{matrix} 1 & 5 \\ 2 \\ 3 \\ 4 \end{matrix}$$

But the remaining boxes (which need to be filled with 6, 7, 8, 9) form what's called a skew tableaux, not a standard tableau. Here it's easy to see that there are 5 possibilities, but in general counting skew tableaux is harder.

$$\begin{matrix} & 6 & 7 \\ 8 & 9 \end{matrix}, \quad \begin{matrix} & 6 & 8 \\ 7 & 9 \end{matrix}, \quad \begin{matrix} & 6 & 9 \\ 7 & 8 \end{matrix}, \quad \begin{matrix} & 7 & 8 \\ 6 & 9 \end{matrix}, \quad \begin{matrix} & 7 & 9 \\ 6 & 8 \end{matrix}. $$

Here's the diagram for the number of standard Young tableaux of shape $(4,3,1,1)$ by the position of the 5 (the counts again sum to 216).

$$\begin{matrix} 0 & 5 & 60 & 18 \\ 6 & 60 & 0 \\ 52 \\ 15 \end{matrix}$$

Answer 2

(A little too long for a comment);

These calculations can be done with online version of SAGE sagecell :

 def hook_formula(mu):
     return factorial(add(k for k in mu))/prod(mu.hook_length(i,j) for i,j in mu.cells())
 a=hook_formula(Partition([4,3,1,1]))
 b=hook_formula(Partition([4,2,1,1]))
 a,b

(answers : 216,90)

See tutorial here