A= {a1,a2...an} 1+2+...+n=a
B={b1, b2, ....,bm} 1+2+...+m=b
f1 : a1, b1),(a2,b2) ...(an,bm) - b cupluri
f2; (a2,b1) (a2,b2)...(a2,bm)= b cupluri
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fn: (an,b1), (a2b2)...(an, bm) b cupluri
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avem b*b *...*b de a ori => card {f1...f n}= b^a