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 correct option is 2.

Let the three-digit number in the first row be represented as a three-digit integer $x$. According to the problem, the number formed in the second row is $2x$, and the number formed in the third row is $3x$.
Since the digits 1 to 9 are arranged in these three rows, each of the three numbers $x$, $2x$, and $3x$ must be a three-digit number. Furthermore, there can be no repetition of digits across all nine positions, meaning the set of nine digits across $x$, $2x$, and $3x$ must be exactly the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 in some order.

First, we determine the range of possible values for $x$:
Since $3x$ must be a three-digit number, the maximum value of $3x$ is 999. This implies:
$x\le 333$
Also, since $x$ is a three-digit number, $x\ge 100$. Since no digits can be repeated and 0 is not allowed, the digits of $x$ must be distinct and non-zero.

The problem states that only three of the four digits 2, 3, 7, and 9 are allowed to be used in the first row (the number $x$). Let us look at the possible three-digit combinations of these digits that can form $x$:
The subsets of three digits chosen from {2, 3, 7, 9} are:
1) {2, 3, 7}
2) {2, 3, 9}
3) {2, 7, 9}
4) {3, 7, 9}

Since $x\le 333$, the hundreds digit of $x$ can only be 1, 2, or 3.
Let us analyze the possible numbers $x$ formed by arranging the three chosen digits from each subset such that the value of the number is less than or equal to 333:

Case 1: Using digits from {2, 3, 7}
Since the hundreds digit must be 2 or 3 (as 7 is too large), the possible values for $x$ are:
- If the hundreds digit is 2, the remaining digits are 3 and 7, giving $x=237$ or $x=273$.
- If the hundreds digit is 3, the remaining digits are 2 and 7, giving $x=327$ (since 372 > 333, we cannot use 7 as the tens digit).
Let's test these values:
- For $x=237$:
$2x=474$ (digit 4 is repeated, not allowed).
- For $x=273$:
$2x=546$
$3x=819$
Let's check the digits used in $x=273$, $2x=546$, and $3x=819$:
Digits: {2, 7, 3, 5, 4, 6, 8, 1, 9}. All digits from 1 to 9 are used exactly once. This is a valid combination.
- For $x=327$:
$2x=654$
$3x=981$
Let's check the digits used in $x=327$, $2x=654$, and $3x=981$:
Digits: {3, 2, 7, 6, 5, 4, 9, 8, 1}. All digits from 1 to 9 are used exactly once. This is also a valid combination.

Case 2: Using digits from {2, 3, 9}
Since the hundreds digit must be 2 or 3, the possible values for $x$ are:
- If the hundreds digit is 2, we have $x=239$ or $x=293$.
- If the hundreds digit is 3, we have $x=329$ (since 392 > 333).
Let's test these values:
- For $x=239$:
$2x=478$
$3x=717$ (digit 7 is repeated).
- For $x=293$:
$2x=586$
$3x=879$ (digits 8 and 9 are repeated).
- For $x=329$:
$2x=658$
$3x=987$ (digit 8 is repeated).

Case 3: Using digits from {2, 7, 9}
The hundreds digit must be 2 (as 7 and 9 are greater than 3).
The possible values for $x$ are $x=279$ or $x=297$.
Let's test these values:
- For $x=279$:
$2x=558$ (digit 5 is repeated).
- For $x=297$:
$2x=594$ (digit 9 is repeated).

Case 4: Using digits from {3, 7, 9}
The hundreds digit must be 3 (as 7 and 9 are greater than 3).
The only possible value for $x$ below 333 is $x=327$ (but the set of digits here is {3, 7, 9}, so we cannot form any number because the other digits must be 7 and 9, making the number at least 379, which is greater than 333).

Thus, there are exactly 2 combinations of three rows that satisfy all constraints:
1) First row: 273, Second row: 546, Third row: 819
2) First row: 327, Second row: 654, Third row: 981

Therefore, the total number of such combinations possible is 2.