Question:easy

Let $S = \{(x, y) \mid x, y \in \mathbb{N}, 1 \le x \le 15, 1 \le y \le 20\}$ be a set. Let $\mathcal{R}$ be the equivalence relation on $S$ defined by $(x, y) \mathcal{R} (x', y')$ if and only if $x + y = x' + y'$. Then the number of equivalence classes of $\mathcal{R}$ on $S$ is

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An equivalence class for $(x, y) \mathcal{R} (x', y')$ is determined by the output of the function $f(x, y) = x + y$.
The number of equivalence classes is equal to the size of the range of this function over the domain $S$.
Updated On: Jun 16, 2026
  • 34
  • 35
  • 15
  • 20
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The Correct Option is A

Solution and Explanation

Step 1: Understand how the classes are formed.
Two pairs $(x, y)$ and $(x', y')$ sit in the same class exactly when their coordinate sums $x + y$ are equal. So each distinct value of the sum $s = x + y$ gives one equivalence class. Counting classes means counting how many different sums are possible.

Step 2: Note the ranges.
Here $x$ runs over $1$ to $15$ and $y$ runs over $1$ to $20$, both whole numbers.

Step 3: Find the smallest possible sum.
The least sum is when both are smallest: $x = 1, y = 1$, giving $s = 2$.

Step 4: Find the largest possible sum.
The greatest sum is when both are largest: $x = 15, y = 20$, giving $s = 35$.

Step 5: Check every value in between is reachable.
As $x$ and $y$ are independent integers covering full ranges, every integer sum from $2$ up to $35$ can be made. For example, to reach any $s$ in this range we can keep $y$ free between $1$ and $20$ and pick a matching $x$ between $1$ and $15$.

Step 6: Count the distinct sums.
The sums run over all integers from $2$ to $35$, which is $35 - 2 + 1 = 34$ values. So there are $34$ equivalence classes.
\[ \boxed{34} \]
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