Such Relations, Much Wow!

Let \(A\) be the set of all primes which cannot be expressed as \(6p\pm1\), \(p\in \mathbb{N}\). Let \(R\) be a random relation mapping from \(A\) to itself. The probability that the relation maps all the elements in the domain to all the elements in the co domain is given by \(\frac{a}{b}\), where \(a,b\in \mathbb{N}\) and \(g.c.d.(a,b)=1\). Find \(a+b\)

Details:

The relation is bijective.

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