Q25 / 48 · JEE Advanced 2026 (Paper 1) · Mathematics
Let $S = \{1, 2, 3, \ldots, 10\}$. Consider the set
$X = \{R : R$ is an equivalence relation on the set $S$ such that $R$ has exactly $42$ elements$\}$.
Then the number of elements in $X$ is _____________.
$X = \{R : R$ is an equivalence relation on the set $S$ such that $R$ has exactly $42$ elements$\}$.
Then the number of elements in $X$ is _____________.
Correct answer: 2520
Solution
An equivalence relation on a set $S$ corresponds to a partition of $S$ into disjoint equivalence classes. Let the sizes of these equivalence classes be $a_1, a_2, \ldots, a_k$.
The number of elements in the equivalence relation $R$ is given by the sum of the squares of the sizes of its equivalence classes:
$\sum_{i=1}^k a_i^2 = 42$
Since the total number of elements in $S$ is $10$, we also have:
$\sum_{i=1}^k a_i = 10$
We need to find all possible partitions of $10$ such that the sum of their squares is $42$. Let's check the possible sizes of the largest equivalence class, $a_1$:
Case 1: $a_1 = 6$
$a_1^2 = 36$. The remaining sum of squares is $42 - 36 = 6$, and the remaining sum of elements is $10 - 6 = 4$.
The only way to partition $4$ such that the sum of squares is $6$ is $2, 1, 1$ (since $2^2 + 1^2 + 1^2 = 6$).
Thus, one valid partition is $\{6, 2, 1, 1\}$.
Case 2: $a_1 = 5$
$a_1^2 = 25$. The remaining sum of squares is $42 - 25 = 17$, and the remaining sum of elements is $10 - 5 = 5$.
The only way to partition $5$ such that the sum of squares is $17$ is $4, 1$ (since $4^2 + 1^2 = 17$).
Thus, another valid partition is $\{5, 4, 1\}$.
Case 3: $a_1 \le 4$
The maximum possible sum of squares would be for the partition $\{4, 4, 2\}$, which gives $4^2 + 4^2 + 2^2 = 36 < 42$. No other partitions can yield a sum of $42$.
Now, we calculate the number of ways to form these partitions from the $10$ elements of $S$.
For the partition $\{6, 2, 1, 1\}$:
The number of ways to divide $10$ elements into groups of sizes $6, 2, 1, 1$ is:
$\dfrac{10!}{6! \times 2! \times 1! \times 1! \times 2!} = \dfrac{3628800}{720 \times 2 \times 1 \times 1 \times 2} = \dfrac{3628800}{2880} = 1260$
(Note: We divide by $2!$ because there are two groups of identical size $1$).
For the partition $\{5, 4, 1\}$:
The number of ways to divide $10$ elements into groups of sizes $5, 4, 1$ is:
$\dfrac{10!}{5! \times 4! \times 1!} = \dfrac{3628800}{120 \times 24 \times 1} = \dfrac{3628800}{2880} = 1260$
Total number of equivalence relations in $X$ is:
$1260 + 1260 = 2520$
Answer: $2520$
The number of elements in the equivalence relation $R$ is given by the sum of the squares of the sizes of its equivalence classes:
$\sum_{i=1}^k a_i^2 = 42$
Since the total number of elements in $S$ is $10$, we also have:
$\sum_{i=1}^k a_i = 10$
We need to find all possible partitions of $10$ such that the sum of their squares is $42$. Let's check the possible sizes of the largest equivalence class, $a_1$:
Case 1: $a_1 = 6$
$a_1^2 = 36$. The remaining sum of squares is $42 - 36 = 6$, and the remaining sum of elements is $10 - 6 = 4$.
The only way to partition $4$ such that the sum of squares is $6$ is $2, 1, 1$ (since $2^2 + 1^2 + 1^2 = 6$).
Thus, one valid partition is $\{6, 2, 1, 1\}$.
Case 2: $a_1 = 5$
$a_1^2 = 25$. The remaining sum of squares is $42 - 25 = 17$, and the remaining sum of elements is $10 - 5 = 5$.
The only way to partition $5$ such that the sum of squares is $17$ is $4, 1$ (since $4^2 + 1^2 = 17$).
Thus, another valid partition is $\{5, 4, 1\}$.
Case 3: $a_1 \le 4$
The maximum possible sum of squares would be for the partition $\{4, 4, 2\}$, which gives $4^2 + 4^2 + 2^2 = 36 < 42$. No other partitions can yield a sum of $42$.
Now, we calculate the number of ways to form these partitions from the $10$ elements of $S$.
For the partition $\{6, 2, 1, 1\}$:
The number of ways to divide $10$ elements into groups of sizes $6, 2, 1, 1$ is:
$\dfrac{10!}{6! \times 2! \times 1! \times 1! \times 2!} = \dfrac{3628800}{720 \times 2 \times 1 \times 1 \times 2} = \dfrac{3628800}{2880} = 1260$
(Note: We divide by $2!$ because there are two groups of identical size $1$).
For the partition $\{5, 4, 1\}$:
The number of ways to divide $10$ elements into groups of sizes $5, 4, 1$ is:
$\dfrac{10!}{5! \times 4! \times 1!} = \dfrac{3628800}{120 \times 24 \times 1} = \dfrac{3628800}{2880} = 1260$
Total number of equivalence relations in $X$ is:
$1260 + 1260 = 2520$
Answer: $2520$
Hence, the correct answer is 2520.