I tried

**nops({x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z });**

and *Maple* out** 9.**

If** x = 1, y = 2, z = 3, **we have

**nops({1, 2, 3, 2, 4, 0, -1, 5, 2});**

equal to** 7.**

**Edit.**

I edited my question. The word "list" into "set".

My question means, there is a triple** (x, y, z) = (1, 2, 3) **so that the number of element** **of the set** {****x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z } **is** 7, **not** 9. **Why we can confirm the number of elements of the set** ****{x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z }** is** 9? **And, how must select** **the integer numbers** x, y, z (0 < x <10, 0< y <10, 0< z < 10) **so that the number of elements of the set** ****{x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z }** is 9**?**

I tried

**restart:**

**ListTools[Categorize]:**

**L:=[]:**

**for x from -2 to 5 do**

**for y from -2 to 5 do**

**for z from -2 to 5 do**

**a:=[x,y,z]:**

**b:=[1+x,2+y,-3+z]:**

**c:=[-2+x,3+y,-1+z]:**

**if op(1,a)*op(2,a)*op(3,a)*op(1,b)*op(2,b)*op(3,b)*op(1,c)*op(2,c)*op(3,c)<> 0**

**and op(1,a)<>op(2,a) and op(1,a)<>op(3,a) and op(1,a)<>op(2,b) and op(1,a)<>op(3,b) and op(1,a)<>op(2,c) and op(1,a)<>op(3,c) and op(2,a)<>op(3,a) and op(2,a)<>op(1,b) and **

**op(2,a)<>op(3,b) and **

**op(2,a)<>op(1,c) and **

**op(2,a)<>op(3,c) and **

**op(3,a)<>op(1,b) and **

**op(3,a)<>op(2,b) and **

**op(3,a)<>op(1,c) and**

**op(3,a)<>op(2,c) and**

**op(1,b)<>op(2,b) and**

**op(1,b)<>op(3,b) and**

**op(1,b)<>op(2,c) and**

**op(1,b)<>op(3,c) and**

**op(2,b)<>op(3,b) and**

**op(2,b)<>op(1,c) and**

**op(2,b)<>op(3,c) and**

**op(3,b)<>op(1,c) and**

**op(3,b)<>op(2,c) and**

**op(1,c)<>op(2,c) and**

**op(1,c)<>op(3,c) and op(2,c)<>op(3,c) then L:=[op(L), {a,b,c}] fi; od: od: od: **

**nops(L); **

**L;**

** **