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A combination takes the number of ways to make an ordered list of n elements (n!), shortens the list to exactly x elements ( by dividing this number by (nx)!B)mn ∈ A so that a−b ∈ N(A), as needed p 270, #44 In order for a ∈ Z 36 to be in N(h0i) we must have an divisible by 36 for some n ∈ Z This happens if and only if a is divisible by all of the prime factors of 36, which are 2 and 3 That is, a must be divisible by 6 Hence N(h0i) =116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l " K \ B \ Z g J b e k d b", L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n
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