No guessing, only counting

Newly introduced randomized variants also pose a good question: how much there are different initial setups? Let's calculate!

Note that randomized Classical Chess variants and randomized Croatian Ties variants of the same size has the same number of possible initial setups. In Croatian Ties variants Knights are replaced by Pegasuses, but otherwise number of different pieces and their count is the same. Pawns in all variants are irrelevant, since they can be arranged in only one way. So, the only difference is the order of figures arranged in the first and last rank. Both mirrored and symmetrical flavors of randomized variants has the same number of combinations, so it's enough to calculate only once, then double the result.

First, a very brief introduction to binomial coefficients:

$$c=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!\cdot (n-k)!}$$

where $n\ge k\ge 0$, and $n!=n\cdot (n-1)\cdot (n-2)\cdot ...\cdot 2\cdot 1$.

Our factor $c$ is number of combinations when placing $k$ pieces of the same kind (e.g. 4 Rooks) onto $n$ available fields (e.g. 10 empty fields).

So, we'll start with randomized Classical Chess 14, and with Bishops, since we have to make sure we have the same number of Bishops on light and dark fields. There are 4 Bishops in this variant, so that would be 2 Bishops on light and other 2 Bishops on dark fields. In these variants there are 14 fields at each rank (surprising, I know!), so that gives us 7 light and 7 dark fields. Obviously, each set of light and dark fields is completely separated from each other, and so numbers of possible arrangements of Bishops on different fields are the same:

$$c_{\mathrm{light}}=c_{\mathrm{dark}}=\left(\genfrac{}{}{0ex}{}{7}{2}\right)=\frac{7!}{2!\cdot 5!}=21$$

where $7$ is the number of either light or dark fields; and $2$ is count of Bishops on those fields. So, our factor $c_{\mathrm{light}}$ is number of combinations of 2 Bishops arranged over 7 light fields; and $c_{\mathrm{dark}}$ is the same over dark fields.

Next, we have to arrange 4 Rooks over $14-4=10$ fields, where 14 is size of a variant, and 4 is number of fields already occupied by Bishops:

$$c_{\mathrm{Rooks}}=\left(\genfrac{}{}{0ex}{}{10}{4}\right)=\frac{10!}{4!\cdot 6!}=210$$

Next, we have to arrange 4 Knights over $14-8=6$ fields, where 14 is still size of a variant, and 8 is number of fields already occupied by Bishops and Rooks:

$$c_{\mathrm{Knights}}=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=\frac{6!}{4!\cdot 2!}=15$$

Queen can now pick between two remaining fields (ladies are choosers! 😎):

$$c_{\mathrm{Queen}}=2$$

King will have to be satisfied with whatever field is left empty:

$$c_{\mathrm{King}}=1$$

So, our number of combinations for randomized Classical Chess 14 variants is:

$$c_{\mathrm{14}}=c_{\mathrm{light}}\cdot c_{\mathrm{dark}}\cdot c_{\mathrm{Rooks}}\cdot c_{\mathrm{Knights}}\cdot c_{\mathrm{Queen}}\cdot c_{\mathrm{King}}=\mathrm{2,778,300}$$

that is for either mirrored or symmetrical flavor. If flavor is also randomized choice, total number of initial setups is doubled:

$$c_{\sum \mathrm{14}}=c_{\mathrm{14}}\cdot 2=\mathrm{2,778,300}\cdot 2=\mathrm{5,556,600}$$

both those numbers also apply to randomized Croatian Ties 14 variant(s).

Calculation of number of different initial setups for randomized Classical Chess 20 variant(s) -and, by extension, Croatian Ties 20 variant(s)- is the same as above, but numbers are slightly bigger. Size of a chessboard is now 20, there are 10 light fields, and 10 dark fields, there are 6 Rooks, 6 Knights and 6 Bishops (of which 3 are on light fields, and other 3 on dark ones). Taking all into account gives the number of initial setups as:

$$c_{\mathrm{20}}=\mathrm{2,421,619,200}$$

that is for either mirrored or symmetrical flavor. If flavor is also randomized choice, total number of initial setups is doubled:

$$c_{\sum \mathrm{20}}=c_{\mathrm{20}}\cdot 2=\mathrm{2,421,619,200}\cdot 2=\mathrm{4,843,238,400}$$

both those numbers also apply to randomized Croatian Ties 20 variant(s).

Number of different initial setups for randomized Classical Chess 26 variant(s) is calculated the same as above, but numbers are bigger still. Size of a chessboard is now 26, there are 13 light fields, and 13 dark fields, there are 8 Rooks, 8 Knights and 8 Bishops (of which 4 are on light fields, and other 4 on dark ones). So, number of initial setups is:

$$c_{\mathrm{26}}=\mathrm{2,013,316,519,500}$$

that is for either mirrored or symmetrical flavor. If flavor is also randomized choice, total number of initial setups is doubled:

$$c_{\sum \mathrm{26}}=c_{\mathrm{26}}\cdot 2=\mathrm{2,013,316,519,500}\cdot 2=\mathrm{4,026,633,039,000}$$

both those numbers also apply to randomized Croatian Ties 26 variant(s).

As an interesting tidbit, if we try to calculate number of initial setups for randomized Classical Chess, we get:

$$c_8=\mathrm{2,880}$$

that is for either mirrored or symmetrical flavor. If flavor is also randomized choice, total number of initial setups is doubled:

$$c_{\sum 8}=c_8\cdot 2=\mathrm{2,880}\cdot 2=\mathrm{5,760}$$

Obviously, these numbers are somewhat bigger than Fischer Random Chess would have it (i.e. 960), since it also imposes an additional constraint on how initial setups are generated; namely, "the King must be placed on a square between the Rooks", and does not support symmetrical flavor, only mirrored.