A number of Young French staircases

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Suppose we are building a staircase with 10 blocks flush against a wall: four blocks on the bottom row, three blocks on the second, two blocks on the third row and one block on the top row. Let's assume that we cannot place blocks either (a) floating in mid-air (likely a good assumption) or (b) away from the wall or another block (less physically necessary, but sure, I'll bite).

How many different ways are there of constructing this 10 block, 4 step (or more generally an $n(n+1)/2$ block, $n$ step) staircase under those conditions?

Let's orient ourselves with the wall standing at the $y$-axis and the $x$-axis is the ground. Then our slippy, slidy blocks will form what is known as a Young diagram in French notation for the partition $(n, n-1, n-2, \dots, 3, 2, 1)$ of $n(n+1)/2$. If we are to number the blocks that we place, then they will form a standard Young tableau, where the numbering is non-decreasing when increasing both row- and columnwise. A lot to unpack there, but we will make it through I'm sure.

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When enumerating standard Young tableau, we can avail ourselves of what's known as the hook length formula, which for our case is given by for our partition $\lambda \in \mathbb{N}^d$ as $$F_\lambda = \frac{\left(\sum_{i=1}^d \lambda_i\right)!}{ \prod_{i,j} h(i,j\mid \lambda) },$$ where $$h(i,j \mid \lambda) = (\lambda_i - j) + (d - i) + 1$$ is the hook length for cell $(i,j)$ in the Young diagram for partition $\lambda.$

For our particular staircases, $\lambda_n = (n, n-1, n-2, \dots, 3, 2, 1)$, so we need to just calculate the hook length for each of the $\sum_i \lambda_{n,i} = n(n+1)/2$ cells of the Young diagram and then can apply the hook length formula. Given the shape of our staircases, we see that for $\lambda_n,$ there are $n$ cells with hook length $1,$ namely the top cell on each column (or correspondingly rightmost cell on each row). Moving one cell inward, we see that there are $n-1$ cells with hook length $3,$ and even further there are $n-k$ cells with hook length $(2k+1),$ for $k = 0, 1, \dots, n-1.$

Therefore, applying the hook length formula we get $$F_n = F_{(n,n-1,n-2, \dots, 3,2,1)} = \frac{\left( \frac{n(n+1)}{2} \right)!}{\prod_{k=0}^{n-1} (2k+1)^{(n-k)}}.$$ For $n=4,$ this gives $F_4 = \frac{10!}{1^4\cdot3^3\cdot5^2\cdot7^1} = 768.$ The OEIS entry A005118 contains the general term (offset by $1$), that is, $A005118(n+1) = F_n.$ Seeing as once you get to only a fairly reasonable $n=6$ steps, there are over $1.1$ billions combinations, I guess I will be satisfied with any possible tableau when constructing my next staircase.