Question 9

An equivalence relation $R$ on a set $S$ corresponds to a partition of $S$ into disjoint subsets $S_1, S_2, \dots, S_k$.

The number of elements in $R$ is given by $|R| = \sum_{i=1}^k n_i^2$, where $n_i = |S_i|$ and $\sum_{i=1}^k n_i = |S| = 10$.

We need to find partitions of 10 such that the sum of squares of the part sizes is 42. Case 1: $n_i$ values are $\{6, 2, 1, 1\}$.

Check: $6^2 + 2^2 + 1^2 + 1^2 = 36 + 4 + 1 + 1 = 42$. Sum: $6+2+1+1=10$.

Number of ways to form this partition: $\frac{10!}{6! 2! 1! 1! 2!} = \frac{10 \times 9 \times 8 \times 7}{2 \times 2} = 1260$.

Case 2: $n_i$ values are $\{5, 4, 1\}$.

Check: $5^2 + 4^2 + 1^2 = 25 + 16 + 1 = 42$. Sum: $5+4+1=10$.

Number of ways to form this partition: $\frac{10!}{5! 4! 1!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{24} = 1260$.

Total number of elements in $X = 1260 + 1260 = 2520$.