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If the value of n(Y)+n(Z) is k2, then |k| is .
36
For each element c, count x S with |x-c| 2: for c=1,6: 4; for c=2,3,4,5: 3.
Y: all 6 pairs have the same second element b, needing 6 choices of a with |a-b| 2 — at most 4 exist, so n(Y)=0.
Z: each a gets one image f(a) with |a-f(a)| 2: n(Z)=4 3 3 3 3 4=1296.
n(Y)+n(Z)=1296=362 |k|=36.
Answer: 36
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