Question:
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
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For relations defined by equations, count valid ordered pairs by matching equal outputs from both sides.
Updated On: Mar 25, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Listing possible values of $2a+3b$.
For $a,b \in \{1,2,3,4\}$, possible values of $2a+3b$ are computed systematically.
Step 2: Listing possible values of $3c+4d$.
For $c,d \in \{1,2,3,4\}$, compute all values of $3c+4d$.
Step 3: Matching equal values.
For every common value obtained in both expressions, count the ordered pairs $((a,b),(c,d))$.
Total number of such matching ordered pairs is \[ 18 \]
For $a,b \in \{1,2,3,4\}$, possible values of $2a+3b$ are computed systematically.
Step 2: Listing possible values of $3c+4d$.
For $c,d \in \{1,2,3,4\}$, compute all values of $3c+4d$.
Step 3: Matching equal values.
For every common value obtained in both expressions, count the ordered pairs $((a,b),(c,d))$.
Total number of such matching ordered pairs is \[ 18 \]
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