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

We need $2a+3b = 3c+4d$ where $a,b,c,d \in \{1,2,3,4\}$.
Possible values of $2a+3b$: range from $5$ to $20$.
Possible values of $3c+4d$: range from $7$ to $28$.
For each common value $k$, the number of elements in $R$ equals (number of $(a,b)$ with $2a+3b=k$) $\times$ (number of $(c,d)$ with $3c+4d=k$).
$k=7$: $(a,b)=(2,1)$, $(c,d)=(1,1)$ $\Rightarrow 1 \times 1=1$
$k=10$: $(2,2)$, $(2,1)$ $\Rightarrow 1$
$k=11$: $(1,3),(4,1)$, $(1,2)$ $\Rightarrow 2$
$k=13$: $(2,3)$, $(3,1)$ $\Rightarrow 1$
$k=14$: $(1,4),(4,2)$, $(2,2)$ $\Rightarrow 2$
$k=15$: $(3,3)$, $(1,3)$ $\Rightarrow 1$
$k=16$: $(2,4)$, $(4,1)$ $\Rightarrow 1$
$k=17$: $(4,3)$, $(3,2)$ $\Rightarrow 1$
$k=18$: $(3,4)$, $(2,3)$ $\Rightarrow 1$
$k=20$: $(4,4)$, $(4,2)$ $\Rightarrow 1$
Total $= 1+1+2+1+2+1+1+1+1+1 = 12$