假设 L = {p, q, r, s, t} 是由以下哈斯图表示的格子：
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对于任何x,y∈L，不一定不同，x∨y和x∧y分别是x,y的连接和相遇。令 L 3 = {(x,y,z): x, y, z ∈ L} 是 L 元素的所有有序三元组的集合。令 pr 是元素 (x,y,z) ∈所选择的 L 3等价地满足 x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)。然后
(A) Pr = 0
(B) Pr = 1
(C) 0 < Pr ≤ 1/5
(D) 1/5 < Pr < 1答案： (D)
解释：
Number of triplets in L3 = Number of ways in which
we can choose 3 elements
from 5 with repetition
= 5 * 5 * 5
= 125.
Now, when we take x = t, then the given condition for L is
satisfied for any y and z. Here, y and z can be taken in
5 * 5 = 25 ways.
Take x = r, y = p, z = p. Here also, the given condition is
satisfied. So, pr > 25 / 125 > 1/5.
For x = q, y = r, z = s, the given condition is not satisfied
as q ⋁ (r ⋀ s) = q ⋁ p = q, while (q ⋁ r) ⋀ (q ⋁ s) = t ⋀ t = t.
So, pr ≠ 1.
Hence D choice.
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