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?

The problem requires finding a 3-digit number (let's call it N) such that:

1. N, 2N, and 3N are all 3-digit numbers.
2. All nine digits used across N, 2N, and 3N must be distinct and from 1 to 9 (no repetition, no 0s).
3. The three digits forming N must be a subset of three digits chosen from {2, 3, 7, 9}.

Let's analyze the conditions:

1. Range for N: Since 3N must be a 3-digit number, 3N = 100. Thus, 100 333), etc., do not work.

Case 2: Digits for N are {2, 3, 9}

• N = 239: 2N = 478, 3N = 717 (digit 7 repeated). Invalid.
• N = 293: 2N = 586, 3N = 879 (digits 8 and 9 repeated). Invalid.
• N = 329: 2N = 658, 3N = 987 (digits 8 and 9 repeated). Invalid.
• Other permutations like 392 (N > 333) do not work.

Case 3: Digits for N are {2, 7, 9}

• N = 279: 2N = 558 (digit 5 repeated). Invalid.
• N = 297: 2N = 594 (digit 9 repeated). Invalid.
• Other permutations like 729 (N > 333) do not work.

Case 4: Digits for N are {3, 7, 9}

• All permutations of these digits (e.g., 379, 397) result in N > 333. Invalid.

Only two combinations satisfy all the conditions: N=273 and N=327.

The final answer is C) 2.

The final answer is C.