Is the ans to this question : 1/9

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.
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Sir,

we have to find the probability that the number of numbers that will be divisible by 9. Wont we have to find the number of numbers to find the actual probability?

1/9 of positive integers are divisible by 9. Unless we insert some sort of bias to our set of possible 7-digit numbers that impacts divisibility by 9, our odds would be 1/9. The only digit we aren't allowed to use is 0. A number is divisible by 9 if the sum of the digits is divisible by 9. The absence of 0 from our universe does not create a bias, so we still have 1/9.

Answer choice C.


Worth noting here that removing 0 WOULD have an impact on divisibility by 2, 4, 5, 6, and 8. But not 9.
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It's only sometimes true that you need to count all of the possibilities to answer a probability question. For example, if you were asked "if a random five-digit number is created using each of the digits 1, 3, 5, 7 and 9 once each, what is the probability the number will be odd?" there's no need to calculate how many five-digit numbers can be made. The result will be odd no matter what, so the probability is 1.



This justification of the answer isn't right, as you'll see if you try to apply the same reasoning to the same question, but where we're making 6-digit numbers instead of 7-digit numbers. The probability a random 6-digit number made from six of the digits from 1 through 9 is divisible by 9 is not 1/9 (it can't be, because we have 9C3 = 84 choices for the digits, and 84 is not a multiple of 9, so we can't get a denominator of 9). The answer to that question turns out to be 5/42 if my quick arithmetic is right.
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5/42 = 0.1190
1/9 = 0.1111

I'll stick with my approach unless GMAC puts answer choices that are that close. Yeah, it's an approximation (I notice you chose to shrink the universe rather than expand it, and the more it's shrunk, the higher the risk of introducing the bias I mention), but if you haven't guessed by now, I believe strongly in ballparking and other approaches that get the right answer without worrying about doing that "quick arithmetic."
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