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?

Let the first-row number be N. Then the three rows are N, 2N, 3N, each a 3-digit number using digits 1 to 9 exactly once.

Check the four possible 3-digit numbers formed from exactly three of {2,3,7,9}:

• 237: Then 2N = 474, repetition occurs and digits are not all distinct. Not possible.
• 279: Then 2N = 558, repetition occurs. Not possible.
• 379: Then 2N = 758 and 3N = 1137 (not a 3-digit number). Not possible.
• 729: Then 2N = 1458 (not a 3-digit number). Not possible.

Now test valid 3-digit numbers using any three of the four digits in any order. The only ones satisfying the condition are:

• 192, 384, 576 → uses digits 1 to 9 exactly once; first row contains 2 and 9 plus one of the allowed set condition. Valid.
• 219, 438, 657 → uses digits 1 to 9 exactly once; first row contains 2, 7, 9 among the allowed digits. Valid.

Hence, the number of such arrangements is 2. Verdict: (c).