# Question:why the length of this set is always equal to 9?

## Question:why the length of this set is always equal to 9?

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;

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