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: Use the property of equivalence relations.
If equivalence classes have sizes: \[ n_1,n_2,\ldots,n_k \] then number of ordered pairs in the equivalence relation is: \[ n_1^2+n_2^2+\cdots+n_k^2 \] Given: \[ n_1+n_2+\cdots+n_k=10 \] and: \[ n_1^2+n_2^2+\cdots+n_k^2=42 \]

Step 2: Find possible class sizes.
We need integers whose sum is: \[ 10 \] and sum of squares is: \[ 42 \] Observe: \[ 5^2+4^2+1^2 = 25+16+1 = 42 \] Also: \[ 5+4+1=10 \] Hence equivalence classes must have sizes: \[ 5,\ 4,\ 1 \]

Step 3: Count the number of such partitions.
Choose: \[ 1 \] element for singleton class: \[ \binom{10}{1} \] From remaining: \[ 9 \] elements, choose: \[ 5 \] elements for the class of size \(5\): \[ \binom{9}{5} \] Remaining: \[ 4 \] elements form the last class automatically. Thus total number: \[ \binom{10}{1}\binom{9}{5} \] \[ =10\times126 \] \[ =1260 \]

Step 4: Identify the final answer.
Therefore: \[ \boxed{1260} \]