Expand Kirkman's Schoolgirl Problem
Mathematica, 935 bytes

Inp={5,4};L=Length;T=Table;ST[t_,k_,n_]:=Binomial[n-1,t-1]/Binomial[k-1,t-1];H=ToExpression@Alphabet[];Lo=Inp[[1]]*Inp[[2]];H=H[[;;Lo]];Final={};ST[2,3,12]=4;ST[2,4,20]=5;If[Inp[[2]]==1,Column[Partition[H,{1}]],CA=Lo*Floor@ST[2,Inp[[2]],Lo];While[L@Flatten@Final!=CA,Final={};uu=0;S=Normal[Association[T[ToRules[H[[Z]]==Prime[Z]],{Z,L@H}]]];PA=Union[Sort/@Permutations[H,{Inp[[2]]}]];PT=Partition[H,Inp[[2]]];While[L@PA!=0,AppendTo[Final,PT];Test=Flatten@T[Times@@@Subsets[PT[[X]],{2}]/.S,{X, L@PT}];POK=T[Times@@@Subsets[PA[[Y]],{2}]/.S,{Y,L@PA}];Fin=Select[POK,L@Intersection[Test,#]==0&amp;];Facfin=T[FactorInteger[Fin[[V]]],{V,L@Fin}];end=T[Union@Flatten@T[First/@#[[W]],{W,L@#}]&amp;[Facfin[[F]]],{F,L@Facfin}]/.Map[Reverse,S];PA=end;PT=DeleteDuplicates[RandomSample@end,Intersection@##=!={}&amp;];If[L@Flatten@PT&lt;L@H,While[uu&lt;1000,PT=DeleteDuplicates[RandomSample@end,Intersection@##=!={}&amp;];If[L@Flatten@PT==L@H,Break[],uu++]]]]];Grid@Final]



this is for 26 ladies max 

EDIT
I made some changes and I think it works!
The code right now is set to solve [5,4] (which is the "social golfers problem") and gets the result in a few seconds.
However [5,3] problem is tougher and you will have to wait 10-20 minutes but you will get a right combination for all days.
For easier cases it is very quick.

anyway you can try it and see the results
Try it online here
copy and paste using ctrl-v
press shift+enter to run the code
you can change the input at the begining of the code -> Inp={5,4}
run the code multiple times to get different permutations