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?

We can only use three of the four digits – 2, 3, 7, and 9, in the first row.

Also, since third row is also a 3 digit number and thrice the first row number, maximum value of first row can be 987/3 ie. 329. So, we are left with following combinations only: 237, 273, 239, 293, 279, 297, 327, and 329.

We will check these numbers by doubling and tripling respectively to check feasibility –

237 × 2 = 474 (digit repetition, so eliminated)
273 × 2 = 546; 273 × 3 = 819 (passes both)
239 × 2 = 478; 239 × 3 = 717 (digit repetition, so eliminated)
293 × 2 = 586; 293 × 3 = 879 (digit repetition, so eliminated)
279 × 2 = 558 (digit repetition, so eliminated)
297 × 2 = 594 (digit repetition, so eliminated)
327 × 2 = 654; 327 × 3 = 981 (passes both)
329× 2 = 658; 329× 3 = 987 (digit repetition, and so eliminated)