Right answer is (a) number of relations
Explanation: For a set with k elements the number of binary relations should be 2^(n*n) and the number of functions should be n^n. Now, 2^(n*n) => n^2log (2) [taking log] and n^n => nlog (n) [taking log]. It is known that n^2log (2) > nlog (n). Hence, the number of binary relations > the number of functions i.e, Nr > Nf.