Permutations and Combinations

Concept:An equivalence relation on a set corresponds to a partition into disjoint equivalence classes, where each class of size $n$ contributes $n^2$ ordered pairs to the relation.Explanation:We have the set $S = \{1,2,3,\ldots,10\}$ and the relation must contain exactly $42$ ordered pairs.If the sizes of the equivalence classes are $n_1, n_2, \ldots, n_k$, then $\sum n_i = 10$ and $\sum n_i^2 = 42$.Find all sets of positive integers satisfying these two conditions.The possible partitions are: $\{5,4,1\}$ and $\{6,2,1,1\}$ because $5^2+4^2+1^2=25+16+1=42$ and $6^2+2^2+1^2+1^2=36+4+1+1=42$.Now count the number of equivalence relations for each partition.Case 1: $\{5,4,1\}$Choose 5 elements for the class of size 5: $\binom{10}{5}$ ways.From the remaining 5 elements, choose 4 for the class of size 4: $\binom{5}{4}$ ways.The last element forms the class of size 1: $\binom{1}{1}=1$ way.Total for this case: $\binom{10}{5} \times \binom{5}{4} = 252 \times 5 = 1260$.Case 2: $\{6,2,1,1\}$Choose 6 elements for the class of size 6: $\binom{10}{6}$ ways.From the remaining 4 elements, choose 2 for the class of size 2: $\binom{4}{2}$ ways.The remaining 2 elements automatically become two separate classes of size 1 each.Since the two classes of size 1 are indistinguishable, we must divide by $2!$ to avoid overcounting.Number of ways: $\binom{10}{6} \times \binom{4}{2} \times \frac{1}{2!} = 210 \times 6 \times \frac{1}{2} = 1260$.Total number of equivalence relations = sum of both cases: $1260 + 1260 = 2520$.Answer:$2520$