Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in

A number will be divisible by 9 if the sum of its digits is divisible by 9. So we aren't concerned with order at all here, because the sum of the digits of our seven-digit number will be the same no matter what order we write the digits in. We just care about what digits we choose.

The nine digits we're starting with sum to 45, so when we leave out two of the digits to make our seven-digit number, for our sum to remain a multiple of 9, the two digits we leave out will need to sum to 9 (they can't sum to 18 or any larger multiple of 9, because the digits aren't big enough). So we'll get a collection of digits that sum to 36 if we leave out the pairs of digits 1, 8, or 2, 7, or 3, 6, or 4, 5, and there are 4 sets of digits we can choose to omit that sum to a multiple of 9. There are 9C2 = 36 pairs of digits in total we could leave out (or equivalently, 9C7 = 36 sets of seven digits we can pick), so the answer is 4/36 = 1/9.

If you were obligated to guess here, then you could notice that we can make 9C7 = 9C2 = 36 seven-digit numbers in total. So it must be possible to write the answer in the form x/36, where x is an integer. When we cancel that fraction down, the denominator will become some divisor of 36. So A, B and E are absolutely impossible, and if you then are choosing between 1/4 and 1/9, since a random number has a 1/9 probability of being divisible by 9, the answer 1/4 seems much too large here (since we're almost picking a random number) and 1/9 is a very reliable guess.