You mean 13-5 does not equal 9?
. Here's the correction.
For the nut flush, villain has to have the king of clubs. Given he has at least two clubs and there is no other information than what OP states, we have:
Total unknown cards: 43. Remaining clubs: 8. Remaining non-clubs 35
The probabilities are:
V has two clubs, Kc Xc : C(1,1)*C(7,1)*C(35,2)/C(43,4) =3.37%
V has three clubs, Kc Xc Yc: C(1,1)*C(7,2)*C(35,1)/C(43,4) =0.60%
V has four clubs: Kc Xc Yc Zc: C(1,1)*C7,3)*C(35,0) = 0.03%
Total =4.00%