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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 03 Jan 2022, 07:45
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order. The probability that this seven digit number will be divisible by 9 is:

(A) \(\frac{1}{45}\)

(B) \(\frac{5}{63}\)

(C) \(\frac{1}{9}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{5}{8}\)


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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 03 Jan 2022, 09:58
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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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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 03 Jan 2022, 09:20
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Is the ans to this question : 1/9
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 25 May 2022, 09:03
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IanStewart
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?
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 25 May 2022, 09:27
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Bunuel
Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order. The probability that this seven digit number will be divisible by 9 is:

(A) \(\frac{1}{45}\)

(B) \(\frac{5}{63}\)

(C) \(\frac{1}{9}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{5}{8}\)


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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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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 25 May 2022, 15:11
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AashishGautam

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?
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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.

ThatDudeKnows

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.
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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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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 25 May 2022, 15:23
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IanStewart
AashishGautam

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?

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.

ThatDudeKnows

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.

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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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 13 Aug 2023, 23:23
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Total sum = 45, we have to remove no such that sum remains divisible by 9.

Eliminate 2 no such that sum= 9,(l,8), (2,7) ' (3,6),4, 5). There fore 4 cases next no is 18 but max is 9+8 -17 so we can't go beyond that probability - 4/(9 x 8 X7 x6 x5 X 4 X 3)]x 7!= 1/9
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 27 Nov 2025, 07:08
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Bunuel
Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order. The probability that this seven digit number will be divisible by 9 is:

(A) \(\frac{1}{45}\)

(B) \(\frac{5}{63}\)

(C) \(\frac{1}{9}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{5}{8}\)


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I believe the easiest way to solve this is to add up all the numbers and then see which two you can subtract,.

Adding up all these numbers, you get 45

Now, the closest number to this that is divisible by 9 is 36

So, this 9 can be (1,8) (2,7) (3,6) (4,5)....we are taking out 2 numbers from total 9 numbers so that we have seven numbers.

Find out total pairs 9C2 = 36

Answer: 4/36 = 1/9
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 28 Nov 2025, 07:20
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Bunuel
Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order. The probability that this seven digit number will be divisible by 9 is:

(A) \(\frac{1}{45}\)

(B) \(\frac{5}{63}\)

(C) \(\frac{1}{9}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{5}{8}\)


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To find the probability that the 7-digit number is divisible by 9, we use the divisibility rule for 9.

The Logic: The "Excluded Digits" Shortcut
Rule: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

1. Total Sum: The sum of the digits from 1 to 9 is:
\(1 + 2 + 3 + ... + 9 = 45\)

2. The Shortcut: We are choosing 7 digits. This is mathematically equivalent to excluding 2 digits.
Let \(S\) be the sum of the 7 chosen digits.
Let \(R\) be the sum of the 2 excluded digits.
\(S + R = 45\)

Since 45 is divisible by 9, for \(S\) to be divisible by 9, \(R\) (the sum of the missing digits) must also be divisible by 9.

***

Calculation

Step 1: Total Possible Pairs of Excluded Digits
We are choosing 2 distinct digits out of 9 to exclude.
\(Total = \binom{9}{2} = \frac{9 \times 8}{2} = 36\) possible pairs.

Step 2: Favorable Pairs (Sum is a multiple of 9)
The maximum sum of two distinct digits is \(8 + 9 = 17\). Therefore, the only multiple of 9 that two digits can sum to is 9 (18 is impossible).

Let's list the pairs that sum to 9:
1. \(\{1, 8\}\)
2. \(\{2, 7\}\)
3. \(\{3, 6\}\)
4. \(\{4, 5\}\)

There are 4 favorable pairs.

Step 3: Probability
\(P = \frac{\text{Favorable Pairs}}{\text{Total Pairs}} = \frac{4}{36}\)

Simplify the fraction:
\(P = \frac{1}{9}\)

This matches Option (C).

Answer: C
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Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in 12 Apr 2026, 12:58
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Another way to think about this exercise—although there are better solutions, in my opinion.

We know that the sum of the digits must be a multiple of 9. Since the minimum sum of the digits 1, 2, ..., 7 is 28 and for 3, 4, ..., 9 is 42, the sum of the digits must be 36 (since it cannot be 27 or 45, as shown).

The next step is to determine how many 7-digit numbers have a digit sum of 36. The way I approached this was somewhat manual and took about 1.1 minutes, which is why I don’t recommend this method.

First, I found 9876321, then 9875421. Then I realized that I must remove 9 from the pairs (5,4), (6,3), (7,2), or (8,1).

With that, the solution is:

7A7 times 4 divided by 9A7 = (7!/0!)*4/(9!/2!) = 7!*8 / 9*8*7! = 1/9
Bunuel
Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order. The probability that this seven digit number will be divisible by 9 is:

(A) \(\frac{1}{45}\)

(B) \(\frac{5}{63}\)

(C) \(\frac{1}{9}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{5}{8}\)


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