I'm not sure if it will work, but maybe try with portals.
Create two portals each with only one row and put your color field in both. Sort the first portal in ascending order so that it gets your first color record. Sort the second portal in descending order so that it gets the other color record.
If you have a lot of different color options this mite not be feasible, but if it is only two then it could be possible.
If each chair can have 3 or more colors you mite need to create additional fields to sort with (not quite sure what that would look like)
Hope this works
To get the two colours, first add a calculation field (two_colours) to your table and check the "do not store calculation results…" checkbox in the storage options.
The calculation is:
Colour & If (Get(FoundCount) > Get(RecordNumber) ; " and " & GetNthRecord(Colour;Get(RecordNumber)+ 1) ; "")
This will result in "red and green" or just "green" if it is the last record.
On the layout, add a merge field (i.e. <<two_colours>> )
P.S.: If you want to drop the plural ("s") from the word "colours" for the last record, then you can use the following calculation instead:
Let ([X = Get(RecordNumber) ; Y = Get(FoundCount)] ; If(Y > X; "s "; " ") & Colour & If (Y > X ; " and " & GetNthRecord(Colour;X+ 1) ; ""))
On the layout, insert this:
The chair is available in the colour<<two_colours>>
Thanks to the both of you. Both solution work, the first only when there's an internal relation in the table. I applied the solution with the calculating field.