Let R be a relation defined on the set {1 , 2, 3, 4} x { l, 2, 3, 4} by R = {((a, b), (c, d)): 2a + 3b = 3c + 4d}.
Then the number of elements in R is

The given relation $$R$$ is defined on the set $$\{1,2,3,4\}\times\{1,2,3,4\}$$.
An element of $$R$$ has the form $$((a,b),(c,d))$$ where $$a,b,c,d\in\{1,2,3,4\}$$ and the condition is

$$2a+3b = 3c+4d \; -(1)$$

Thus the task is to count all ordered quadruples $$(a,b,c,d)$$ satisfying $$-(1)$$.

Step 1: List all values of $$2a+3b$$

Compute $$2a+3b$$ for every $$a,b\in\{1,2,3,4\}$$ (there are $$4\times4=16$$ combinations).
  • For $$b=1$$: $$2a+3 = 5,7,9,11$$ when $$a=1,2,3,4$$ respectively.
  • For $$b=2$$: $$2a+6 = 8,10,12,14$$ when $$a=1,2,3,4$$.
  • For $$b=3$$: $$2a+9 = 11,13,15,17$$ when $$a=1,2,3,4$$.
  • For $$b=4$$: $$2a+12 = 14,16,18,20$$ when $$a=1,2,3,4$$.

Count the frequency $$L(t)$$ of each value $$t$$:

$$\begin{array}{c|c} t & L(t)\\\hline 5 & 1\\ 7 & 1\\ 8 & 1\\ 9 & 1\\ 10 & 1\\ 11 & 2\\ 12 & 1\\ 13 & 1\\ 14 & 2\\ 15 & 1\\ 16 & 1\\ 17 & 1\\ 18 & 1\\ 20 & 1 \end{array}$$

Step 2: List all values of $$3c+4d$$

Compute $$3c+4d$$ for every $$c,d\in\{1,2,3,4\}$$ (again $$16$$ combinations).
  • For $$d=1$$: $$4+3c = 7,10,13,16$$ when $$c=1,2,3,4$$.
  • For $$d=2$$: $$8+3c = 11,14,17,20$$ when $$c=1,2,3,4$$.
  • For $$d=3$$: $$12+3c = 15,18,21,24$$ when $$c=1,2,3,4$$.
  • For $$d=4$$: $$16+3c = 19,22,25,28$$ when $$c=1,2,3,4$$.

Count the frequency $$R(t)$$ of each value $$t$$ that appears up to $$t=20$$ (larger values will not match any $$2a+3b$$):

$$\begin{array}{c|c} t & R(t)\\\hline 7 & 1\\ 10 & 1\\ 11 & 1\\ 13 & 1\\ 14 & 1\\ 15 & 1\\ 16 & 1\\ 17 & 1\\ 18 & 1\\ 20 & 1 \end{array}$$

Step 3: Count common solutions

For a fixed value $$t$$, every left triple that gives $$t$$ can pair with every right triple that gives the same $$t$$. Hence the total number of solutions is

$$\displaystyle N = \sum_{t} L(t)\,R(t)$$

Compute the sum over the common values $$t$$ (those that appear in both tables):

$$\begin{aligned} t=7&: L(7)\,R(7)=1\cdot1=1\\ t=10&: 1\cdot1=1\\ t=11&: 2\cdot1=2\\ t=13&: 1\cdot1=1\\ t=14&: 2\cdot1=2\\ t=15&: 1\cdot1=1\\ t=16&: 1\cdot1=1\\ t=17&: 1\cdot1=1\\ t=18&: 1\cdot1=1\\ t=20&: 1\cdot1=1 \end{aligned}$$

Add them: $$1+1+2+1+2+1+1+1+1+1 = 12$$.

Final answer: The relation $$R$$ contains $$12$$ ordered pairs.
Hence the correct option is Option C.