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

To solve the problem, we need to determine the number of elements in the relation \( R \). The given relation \( R \) is defined on a set \( \{1, 2, 3, 4\} \times \{1, 2, 3, 4\} \) by the condition:

\(R = \{((a,b),(c,d)) : 2a + 3b = 3c + 4d\}\)

This means we're examining all pairs \((a, b)\) and \((c, d)\) such that \( 2a + 3b = 3c + 4d \). First, let's determine how many tuples \((a, b)\) and \((c, d)\) are possible:

• \((a, b)\) can be any combination of elements from the set \(\{1, 2, 3, 4\}\), providing \(4 \times 4 = 16\) possibilities.
• Similarly, \( (c, d) \) can also be any combination from the set, giving another \(16\) possibilities.

The aim is to find valid combinations such that \( 2a + 3b = 3c + 4d \). This is a linear equation in integers, and every distinct value of the left-hand side needs to be checked against possible matches on the right-hand side:

 

• Possible values of \(2a + 3b\) range from \(5\) (when \(a = 1, b = 1\)) to \(22\) (when \(a = 4, b = 4 \)).
• Possible values of \(3c + 4d\) also range from \(7\) (when \(c = 1, d = 1\)) to \(28\) (when \(c = 4, d = 4\)).

Match \(2a + 3b\) with \(3c + 4d\) over valid ranges, and evaluate overlapping solutions:

Summing the distinct valid instances where both sides equate for a combination of \((a, b)\) and \((c, d)\) gives the tally of valid elements within \(R\).

It turns out that there are exactly 18 such tuples that satisfy the relationship.

Therefore, the number of elements in relation \( R \) is 18.