(A) 6

(B) 15

(C) 12

(D) 18

Correct Answer: 12

Explanation

Here $a,b,c,d \in \{1,2,3,4\}$. First, list all possible values of $2a+3b$.

For $(a,b)$ from the given set:

$$ \begin{aligned} (1,1)&:5,\quad (1,2):8,\quad (1,3):11,\quad (1,4):14\\ (2,1)&:7,\quad (2,2):10,\quad (2,3):13,\quad (2,4):16\\ (3,1)&:9,\quad (3,2):12,\quad (3,3):15,\quad (3,4):18\\ (4,1)&:11,\quad (4,2):14,\quad (4,3):17,\quad (4,4):20 \end{aligned} $$

Now list all possible values of $3c+4d$.

$$ \begin{aligned} (1,1)&:7,\quad (1,2):11,\quad (1,3):15,\quad (1,4):19\\ (2,1)&:10,\quad (2,2):14,\quad (2,3):18,\quad (2,4):22\\ (3,1)&:13,\quad (3,2):17,\quad (3,3):21,\quad (3,4):25\\ (4,1)&:16,\quad (4,2):20,\quad (4,3):24,\quad (4,4):28 \end{aligned} $$

Common values of $2a+3b$ and $3c+4d$ are:

$$ 7,\;10,\;11,\;13,\;14,\;15,\;16,\;18,\;20 $$

Counting valid ordered pairs $((a,b),(c,d))$ satisfying the equality gives a total of:

$$ \boxed{12} $$

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