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
 

• A. 18
• B. 12
• C. 6
• D. 15

Hint:

For relations defined by equations, count valid ordered pairs by matching equal outputs from both sides.

Answer 1

The Correct Option is A

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 \[ 18 \]