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KEADM
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M27-06 27 Jul 2022, 02:29
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The question says 'n' numbers are selected i.e. they are already selected.
All we need to do is find the probability of one possible arrangement of those numbers (here ascending order) out of total possible arrangements (here n!).
Hence, probability is 1/n!
Opt.B is sufficient.
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M27-06 17 Nov 2022, 12:19
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Hello,

Please review this solution -

1. When any 2 numbers are picked from a set of distinct numbers, say 'a' first and then 'b', the probability of a>b will be 0.5 and probability for a<b is 0.5. So, if n numbers are picked one after another, then the probability in all cases would be. (0.5)^n.

So all we need is the value of 'n'.

Hence, B is correct.
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M27-06 18 Nov 2022, 00:15
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nikhilvsh
Hello,

Please review this solution -

1. When any 2 numbers are picked from a set of distinct numbers, say 'a' first and then 'b', the probability of a>b will be 0.5 and probability for a<b is 0.5. So, if n numbers are picked one after another, then the probability in all cases would be. (0.5)^n.

So all we need is the value of 'n'.

Hence, B is correct.
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Check here: https://gmatclub.com/forum/m27-184482-40.html#p2964973

When you select 2, you can arrange them in 2 ways:
smaller, greater
greater smaller

When you select 3, you can arrange them in 3! ways only one of which is the ascending order.
small, mid, large (ascending order)
mid, small, large
small, large, mid
mid, large, small
large, small, mid
large, mid, small

Similarly, when you select 5, you can arrange them in 5! ways only one of which is in ascending order.
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. To me "ascending order" is ambiguous. As far as I understood, the solution explains that there is only one possible way of selecting the numbers in ascending order. However, wouldn't it also imply that (2,4,7,9,12) would also fall under that category?

Thank you for your help!
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M27-06 01 Mar 2023, 03:03
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andappsc
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. To me "ascending order" is ambiguous. As far as I understood, the solution explains that there is only one possible way of selecting the numbers in ascending order. However, wouldn't it also imply that (2,4,7,9,12) would also fall under that category?

Thank you for your help!
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The ascending order of numbers refers to arranging them from smallest to largest. It's important to note that for any set of five different numbers, there is only one way to arrange them in ascending order. For instance, given the numbers 4, 7, 2, 12, and 9, their ascending order is 2, 4, 7, 9, and 12. Hence, no matter which 5 numbers are selected, the number of ways to select them is always 5!, and only one of those selections will be in ascending order. Therefore, the probability of picking 5 numbers in ascending order is 1/5!.

As this is a very challenging, conceptual question, I recommend carefully reading through the solution and discussion.

Hope it helps.
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M27-06 01 Mar 2023, 22:33
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andappsc
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. To me "ascending order" is ambiguous. As far as I understood, the solution explains that there is only one possible way of selecting the numbers in ascending order. However, wouldn't it also imply that (2,4,7,9,12) would also fall under that category?

Thank you for your help!
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Actually it is a good, conceptual question and takes less than a minute to do - all the characteristics of a GMAT question!
Go through all the explanations provided properly and you may just have an 'Aha' moment! :)
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I think this is a high-quality question and I agree with explanation.
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M27-06 07 Jul 2023, 23:42
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PART-1 -- Any PnC question, start with 2 questions:
1. With / Without replacement?
a. If with replacement --> Possible outcomes = k^n
b. Without replacement --> Possible outcomes can an be kCn / kPn
In this case, it can be (kCn/kPn)

2. Does the order of picking/selection matter? -->
a. If YES --> Possible outcomes = kPn
b. If NO --> Possible outcomes = kCn
In this case, it is kPn

So, Total number of outcomes possible = kPn. This part is now sorted.

PART-2
Out of kPn outcomes, in how many outcomes are the chosen numbers in ascending order?

Let's apply our knowledge a little here before moving to the statements, Out of every n! outcomes (of arranging n numbers), only one outcome will be in ascending order. So for n! outcomes, the probability is --> 1/n!.

What this means is --> Out of n! outcomes, only one is counted as a valid outcome --> Wait a minute! Isn't this how nCr is defined? Think about it, you divide the entire nPr term by n!, to disregard the order of chosen items, you get nCr.

And in this problem as well, we are essentially disregarding the order by choosing only one order out of every n! orders that are possible. Of course, there can be more than one way of choosing 'n' numbers from 'k', we are not discounting that. We are only saying that for all the different sets of 'n' numbers that can be chosen from 'k' numbers, only one in each chosen set will be in ascending order. So all we have to do in kCn.

Therefore, number of favourable outcomes when choosing 'n' numbers from 'k' --> kCn

Probability --> kCn / kPn = 1 / n!

So, this means, the answer DOES NOT depend on the sample space. It only depends on how many selections are made.

So all we need is n, which is given in (2) --> ANS - (B)

Hope this helps, of course, if we had to do this much analysis during the actual exam, this question is NOT a good investment of your time. Please BAIL ASAP!

However, if you came to the exam with this learning already in your kitty, then you earned yourself a brownie point.

This is just my honest attempt at explaining what I think might be a clear explanation. In my view, almost all the questions in GMAT Club tests are of very high quality! And I really use this opportunity to thank Bunuel for that! Let's discuss if you have questions on this, I think this is a BRILLIANT question to clear up our grey areas in PnC.
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M27-06 30 Jul 2023, 19:32
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Hi guys !
Just a quick question at this point. I've read most of the thread, and does my way of thinking, actually works ? On how to reach the 1/5! thing. Especially for people where the 1/5! is not immediate.

So, let's say we have to pick up a set of n numbers, in a k-sized set, without replacement. What are the chances to get a set in an ascending order ?

So, the first pick that I'm doing, gives me 1 chance out of the n (future) picks that my first pick is the smallest.
The second pick is 1 chance out of the (n-1) numbers left. Out of all the numbers that I have left to pick ( -> (n-1) ), only one is the smallest, hence 1/(n-1).
Etc, so at the end, the total chances are:
1/n * 1/(n-1) * 1/(n-2) ....... = 1/n!

Makes sense ?
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M27-06 09 Aug 2023, 05:21
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Total number of ways we can select n numbers from k distinct numbers (ordering matters) = kPn = kCn * n!

Total number of ways we can select n numbers from k distinct numbers such that they are in increasing order = kCn * 1
= number of ways we can select any n numbers
This is so because if we select any n numbers there is only 1 way it can be in increasing order.

So the probability of selecting n number from k numbers such that the numbers are in increasing order = kCn / (kCn * n!)
=1/n!

So only n is required to calculate probability.
Hence B is the answer.
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. In case there are 1000 elements in the Set out which n have to be chosen, then will the probability remail same 1/n!. As I understand that the total no of subset will be 100Cn.
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deepak.brijendra
Bunuel
Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set one-by-one and without replacement, where \(n \le k\), what is the probability that numbers will be selected in ascending order?


(1) Set \(A\) consists of 12 even consecutive integers.

(2) \(n=5\).

It is important to note two things:

1. The probability of selecting any subset of \(n\) numbers from the set is the same. No particular subset of \(n\) numbers is favored over any other subset of \(n\) numbers, so they all have an equal probability of being selected.

2. Any subset of \(n\) distinct numbers can be selected in \(n!\) ways, and only one out of these \(n!\) possible ways will be in ascending order. For example, subset of {1, 2, 3} can be selected in 6 ways: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) or (3, 2, 1). As you can see, out of these \(3!\) possible ways, only one is in ascending order, (1, 2, 3). Therefore, the probability of selecting a subset in ascending order is \(\frac{1}{n!}\).

Thus, to answer the given question, we only need to know the size of the subset (\(n\)) that we are selecting from the set \(A\). Therefore, the firs statement is not sufficient, while the second one is.


Answer: B

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. In case there are 1000 elements in the Set out which n have to be chosen, then will the probability remail same 1/n!. As I understand that the total no of subset will be 100Cn.
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The key point you might be missing is that the question revolves around the order in which the numbers are selected, not which numbers are chosen. The total number of elements in the set is also irrelevant to the probability of selecting numbers in ascending order. No matter if the set has 10, 100, or 1000 elements, if you're selecting n numbers one-by-one, there are n! ways to select them. Only one of these ways will be in ascending order. Hence, the probability remains 1/n! regardless of the size of the set.

I'd advise you to re-read the provided solution carefully. Also, I'd advise you to go through this thread again, because this concept is explained in detail several times here.
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M27-06 02 Feb 2025, 04:05
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I like the solution - it’s helpful. Such sharp question. Thank you for putting this together!
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M27-06 31 Oct 2025, 11:37
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but there are multiple ways of ascending order. Suppose the numbers are 1-12 so 1,3,4,5,6 or 1,2,3,4,5 can be erased. do we have to calculate total ascending orders or like one out of all and hence the probability of 1/120? Bunuel
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Robert wrote \(k\) distinct numbers on an empty blackboard. If David erased \(n\) numbers one by one from the blackboard, where \(n \le k\), what is the probability that the numbers were erased in ascending order?


(1) Robert wrote 12 even consecutive integers on the blackboard.

(2) \(n=5\).
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M27-06 31 Oct 2025, 11:43
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Saun2511
but there are multiple ways of ascending order. Suppose the numbers are 1-12 so 1,3,4,5,6 or 1,2,3,4,5 can be erased. do we have to calculate total ascending orders or like one out of all and hence the probability of 1/120? Bunuel

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You're mixing up which order the problem refers to. The “ascending order” in the question is not about which specific numbers were chosen (like 1,3,4,5,6 vs 1,2,3,4,5). It’s about the sequence of erasing those n numbers once a particular subset is already chosen.

Out of n! possible ways to erase the chosen n numbers, only one sequence (the one where they’re erased from smallest to largest) is in ascending order. That’s why the probability is 1/n!, not something involving how many subsets exist.

Please, go through the previous two pages of discussion in that thread carefully, it should clear up your confusion fully.
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M27-06 09 Nov 2025, 08:39
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How the probability can be found by n=5
we do not know the value of k here so how the probability is found?? as in question probability is asked
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How the probability can be found by n=5
we do not know the value of k here so how the probability is found?? as in question probability is asked
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Please invest more time in reviewing the discussion carefully.
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M27-06 09 Nov 2025, 09:01
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yes i got my answer. thanks sir
Bunuel


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