I change my question to ~~be clearer. ~~clarify.
In the following code (I have integrated the solution for `\tiny`

~~ :~~

):

First define a function:

`def `~~All_pref(cand=["A","B","C","D"],code=1) ~~All_pref(cand=["A", "B", "C", "D"], code=1) :
~~ncand=len(cand)
Scand=sorted(Set(cand))
all_pref=Arrangements(Scand,ncand).list()
all_pref1=[str(Word(x)) ~~ncand = len(cand)
Scand = sorted(Set(cand))
all_pref = Arrangements(Scand, ncand).list()
all_pref1 = [str(Word(x)) for x in all_pref]
if ~~code==1 :
~~code == 1:
return ncand
if ~~code==2 :
~~code == 2:
return ~~all_pref1~~all_pref1

Then use it:

~~cand=["A","B","C","D"]
ne1=[18,16,14,12,11,20,19,14,16,12,2,1,0,0,20,16,13,15,11,10,9,8,7,5]
ne=ne1
Ap=All_pref(cand,2)
cond1=['$\\tiny${}{}{}{}$'.format(*w) ~~cand = ["A", "B", "C", "D"]
ne1 = [18, 16, 14, 12, 11, 20, 19, 14, 16, 12, 2, 1, 0, 0, 20, 16, 13, 15, 11, 10, 9, 8, 7, 5]
ne = ne1
Ap = All_pref(cand,2)
cond1 = ['$\\tiny${}{}{}{}$'.format(*w) for w in Ap]
~~ ~~cond2 ~~=[""]+ ~~= [""] + [r'${}$'.format(*w) for w in cand]
~~ ~~cond3 ~~=[r'${}$'.format(*w) ~~= [r'${}$'.format(*w) for w in cand]
~~ t1=table([ne],header_row=cond1)
rank=[[x.find(l) ~~t1 = table([ne], header_row=cond1)
rank = [[x.find(l) for x in Ap] for l in cand]
~~ score_de_Borda=[[abs(x.find(l)-len(cand)) ~~score_de_Borda = [[abs(x.find(l) - len(cand)) for x in Ap] for l in cand]
~~ t2=table(rank,header_row=cond1,header_column=cond2)
total_score_de_Borda=[add([score_de_Borda[j][i]*ne[i] ~~t2 = table(rank,header_row=cond1,header_column=cond2)
total_score_de_Borda = [add([score_de_Borda[j][i]*ne[i] for i in range(24)]) for j in range(len(cand))]
~~ #t3=table(total_score_de_Borda,header_column=cond3)
~~# t3 = table(total_score_de_Borda,header_column=cond3)
show(cond3,total_score_de_Borda)

~~all ~~everything works fine ~~les ~~apart from the `,header_column=cond3`

(without it I have no problem) and as ~~show(cond3,total_score_de_Borda)~~

shows show(cond3, total_score_de_Borda) shows, the two elements have the same length.