Question

Let \[ S=\{1,2,3,\ldots,10\}. \] Consider the set \[ X=\{R : R \text{ is an equivalence relation on } S \text{ such that } R \text{ has exactly } 42 \text{ elements}\}. \] Then the number of elements in \(X\) is __________.

Hint:

If equivalence classes have sizes: \[ n_1,n_2,\ldots,n_k \] then total elements in the relation are: \[ n_1^2+n_2^2+\cdots+n_k^2 \]

Answer 1

Correct Answer: 1260

Step 1: Understanding the Concept:
An equivalence relation \( R \) on a set \( S \) of size \( n \) is completely defined by a partition of \( S \) into disjoint subsets (equivalence classes) \( S_1, S_2, \dots, S_k \). The number of elements in the relation \( R \) is given by the sum of the squares of the sizes of these classes: \( |R| = \sum_{i=1}^k |S_i|^2 \).
Step 2: Key Formula or Approach:
We need to find a partition of 10 into integers \( n_1, n_2, \dots, n_k \) such that:
1) \( n_1 + n_2 + \dots + n_k = 10 \)
2) \( n_1^2 + n_2^2 + \dots + n_k^2 = 42 \)
Step 3: Detailed Explanation:
Trying possible partitions:
- Let the largest class be \( n_1 = 6 \). Then \( n_1^2 = 36 \). Sum of remaining squares = \( 42 - 36 = 6 \). Remaining elements = \( 10 - 6 = 4 \). Sum of 4 positive integer squares can at most be \( 1^2+1^2+1^2+1^2 = 4 \). No solution.
- Let \( n_1 = 5 \). Then \( n_1^2 = 25 \). Sum of remaining squares = \( 42 - 25 = 17 \). Remaining elements = \( 10 - 5 = 5 \).
If \( n_2 = 4 \), then \( n_2^2 = 16 \). Sum of remaining squares = \( 17 - 16 = 1 \). Remaining elements = \( 5 - 4 = 1 \). This works with \( n_3 = 1 \).
So the partition is \( \{5, 4, 1\} \).
Check: \( 5+4+1 = 10 \) and \( 5^2+4^2+1^2 = 25+16+1 = 42 \).
- Are there other partitions?
If \( n_1 = 4 \), \( n_1^2 = 16 \). Remainder sum of squares = 26, Remainder elements = 6.
If \( n_2 = 4 \), Remainder square sum = 10, Remainder elements = 2. Max square sum is \( 1^2+1^2=2 \). No.
If \( n_2 = 3 \), Remainder square sum = 17, Remainder elements = 3. Max square sum is \( 1^2+1^2+1^2=3 \). No.
So the only partition is \( \{5, 4, 1\} \).
Now we calculate the number of ways to form such classes from 10 distinct elements:
Number of ways = \( \frac{10!}{5! \cdot 4! \cdot 1!} \)
Calculation: \( \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 10 \cdot 9 \cdot 2 \cdot 7 = 1260 \).
Step 4: Final Answer:
The number of such relations is 1260.