The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row; and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows?

Correct Answer: (C) 2

Let the 3-digit number in the first row be $N$. Then the second row is $2N$ and the third row is $3N$, using all digits 1 to 9 without repetition. Testing 3-digit numbers formed by three of the allowed digits $\{2, 3, 7, 9\}$: (1) For $N = 273$, $2N = 546$ and $3N = 819$, which together contain all digits 1 through 9 exactly once. (2) For $N = 327$, $2N = 654$ and $3N = 981$, which also contain all digits 1 through 9 exactly once. No other valid combination exists, giving exactly 2 possible combinations.