We shall use the fact that the number of all subsets of "n"-element set is "2^n."
(a) Let "S=\\{a,b,c,d,e\\}." By defenition, a biniry relation "\\rho" on a set "S" is a subset of a Cartesian square "S\\times S." Since "|S\\times S|=5\\cdot 5=25", there are "2^{25}=33,554,432" binary relations on the set "S=\\{a,b,c,d,e\\}."
(b) Let us find the number of binary relations "\\rho" on the set "S=\\{a,b,c,d,e\\}" that contain "(a, a)" and "(b, b)". Since elements of a relation "\\rho\\setminus\\{(a,a),(b,b)\\}" can be arbitrary elements of "S\\times S\\setminus\\{(a,a),(b,b)\\}", the number of binary relation on "S" that contain "(a, a)" and "(b, b)" is equal to the number of binary relations "R\\subset S\\times S\\setminus\\{(a,a),(b,b)\\}". Since "|S\\times S\\setminus\\{(a,a),(b,b)\\}|=25-2=23" , there are "2^{23}=8,388,608" binary relations on the set "S=\\{a,b,c,d,e\\}" that contain "(a, a)" and "(b, b)".
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