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Re: M2706
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25 May 2019, 07:46
Great question. This question is not about selecting n numbers in ascending order from K numbers. The problem says n numbers from k numbers are already selected. We need to find out what is the probability that these n numbers are in ascending order?



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20 Jul 2019, 14:44
Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B What if, for example, K = 10, and N = 5. aren't both 1,2,3,4,5 and 2,3,4,5,6 in ascending order? So if K=10 and N=5 are the constraints, then there would be 1,2,3,4,5 2,3,4,5,6 3,4,5,6,7 4,5,6,7,8 5,6,7,8,9 6,7,8,9,10 6 ways to select 5 numbers in ascending order. What is wrong with this logic?



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21 Jul 2019, 02:07
tortiees wrote: Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B What if, for example, K = 10, and N = 5. aren't both 1,2,3,4,5 and 2,3,4,5,6 in ascending order? So if K=10 and N=5 are the constraints, then there would be 1,2,3,4,5 2,3,4,5,6 3,4,5,6,7 4,5,6,7,8 5,6,7,8,9 6,7,8,9,10 6 ways to select 5 numbers in ascending order. What is wrong with this logic? That's not what the question asks. We have k distinct numbers. The question asks if we select n numbers from k, what is the probability that those n numbers will be selected in ascending order. k does NOT matter. Say n = 5, and say even though it also does not matter, those 5 numbers are 2, 7, 11, 20, and 100. There are 5! ways to select those numbers and only one out of 5! ways of selecting will be in ascending order, namely {2, 7, 11, 20, 100}. So, P = 1/5!.
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Re: M2706
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21 Jul 2019, 08:11
tortiees wrote: Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B What if, for example, K = 10, and N = 5. aren't both 1,2,3,4,5 and 2,3,4,5,6 in ascending order? So if K=10 and N=5 are the constraints, then there would be 1,2,3,4,5 2,3,4,5,6 3,4,5,6,7 4,5,6,7,8 5,6,7,8,9 6,7,8,9,10 6 ways to select 5 numbers in ascending order. What is wrong with this logic? The essence of the question is you don't need to select 5 numbers. That means whether it is [1,2,3,4,5] or [2,3,4,5,6] is already selected. No need to bother about this selection. Now think, as it is already selected as [2,3,4,5 and 6], in how many ways you can make them ascending? it is invariably 1 and only 1 way.



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27 Jul 2019, 01:25
I think this is a highquality question and the explanation isn't clear enough, please elaborate.



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05 Aug 2019, 11:17
I think this is a highquality question and I don't agree with the explanation. So as per solution you are taking the case where n=k but what about when n<k lets say n=5 and k=6,set A={1,2,3,4,5,6}, now when n=5 probability will be 2 /5! and not 1/5! , because the n=5 can be {1,2,3,4,5} and {2,3,4,5,6}



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06 Aug 2019, 00:26
Anand0802 wrote: I think this is a highquality question and I don't agree with the explanation. So as per solution you are taking the case where n=k but what about when n<k lets say n=5 and k=6,set A={1,2,3,4,5,6}, now when n=5 probability will be 2 /5! and not 1/5! , because the n=5 can be {1,2,3,4,5} and {2,3,4,5,6} The question and the solution are 100% correct. Did you try reading the whole thread? It helps in most cases.
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