As we underlined in Section 1.11, we can distinguish generating non-invertible knots and non-invertible knots that are members of already derived families. From every generating non-invertible pretzel knot its corresponding family is obtained by the following substitutions, respecting symmetry conditions: every single bigon 2 can be replaced by the chain of bigons (2p); chain 3 can be replaced by any odd chain (2p+1); tangle 1 remains unchanged. For n=10 crossings there is only one non-invertible pretzel knot: 2 2,3,2 1. For n=11 there are four of them: 4 1,2 1,3 2 3,2 1,3 2 1 1,2 2,2 1 3 1 1,3,2 1 For the first knot 4 1,2 1,3, its generating knot is 2 1,2 1,3 with n=8 crossings that satisfies the necessary condition for non-invertibility of pretzel knots: its type [0,0,1] consists only from 0-s and 1-s. It contains the tangles (2p) 1 and (2q) 1, so according to the symmetry condition it generates the family of non-invertible pretzel knots (2p) 1,(2q) 1,(2r+1) for p ¹ q. Hence, the first non-invertible knot in this family is 4 1,2 1,3. For n=11, the four non-invertible knots are divided into two subsets: 4 1,2 1,3 2 3,2 1,3 2 1 1,2 2,2 1 3 1 1,3,2 1 The first subset contains the knot 4 1,2 1,3 that belongs to the family of non-invertible knots (2p) 1,(2q) 1,(2r+1) with the additional symmetry condition p ¹ q, and the knots from the other subset generate families of non-invertible knots without additional requirements on parameters.