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Re M2706
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16 Sep 2014, 01:27
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
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Re: M2706
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13 Nov 2014, 07:27
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 Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways please clear the understanding



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Re: M2706
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14 Nov 2014, 02:51
shyam593 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 Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways please clear the understanding Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!.
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Re: M2706
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14 Nov 2014, 05:59
Bunuel wrote: shyam593 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 Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways please clear the understanding Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. ok got it SO we can say Probability = possible outcomes/ total outcomes = 12C5/12P5 12C5: shows selection of 5 numbers and out of which every subset will have 1 set of ascending pairs 12P5: shows selection of 5 numbers and which can be arranged in 5! ways. Cheers!!!!



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Re M2706
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28 Dec 2014, 21:46
Hello,
Here are my thoughts regarding the option B. For instance, let set A = {1,2,3,4,5} where k =5 and n =5.
Probability of selecting the numbers in ascending order = no: of selections in ascending order/ total no: of selections
=> 1/5!
If the set has 6 values  {1,2,3,4,5,6} then, the probability is 2/6.5.4.3.2 which is not 1/6!
probable chances for the above set are 
{1,2,3,4,5} and {2,3,4,5,6}.
Please help me understand why B is correct.



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Re: M2706
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29 Dec 2014, 05:19
vivekb128 wrote: Hello,
Here are my thoughts regarding the option B. For instance, let set A = {1,2,3,4,5} where k =5 and n =5.
Probability of selecting the numbers in ascending order = no: of selections in ascending order/ total no: of selections
=> 1/5!
If the set has 6 values  {1,2,3,4,5,6} then, the probability is 2/6.5.4.3.2 which is not 1/6!
probable chances for the above set are 
{1,2,3,4,5} and {2,3,4,5,6}.
Please help me understand why B is correct. the question does not say consecutive ascending numbers, it just says ascending numbers, so u get wrong ans with k=6 and n=5 try again u will get it



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Re: M2706
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15 Jul 2015, 10:37
vivekb128 wrote: Hello,
Here are my thoughts regarding the option B. For instance, let set A = {1,2,3,4,5} where k =5 and n =5.
Probability of selecting the numbers in ascending order = no: of selections in ascending order/ total no: of selections
=> 1/5!
If the set has 6 values  {1,2,3,4,5,6} then, the probability is 2/6.5.4.3.2 which is not 1/6!
probable chances for the above set are 
{1,2,3,4,5} and {2,3,4,5,6}.
Please help me understand why B is correct. I´m with you though I´d change a bit. If N=5 {1,2,3,4,5} and K=5, then prob should be 1/5! whereas if N=6 {1,2,3,4,5,6} and K=5, probability should be 1/360 (or, 1/6*1/5*1/4*1/3*1/2*2 since my first number can either be 1 or 2). So I´m also not clear how can B be sufficient



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Re: M2706
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29 Jul 2015, 18:03
The only consideration is given to the 5 numbers that you selected. What is the probability that when you select your numbers n1,n2,n3,n4,n5 that n1 < n2 < n3 < n4 < n5 (that is selection 1 is less than selection 2 is less than selection 3...)
There are 5! possible selection orders when you choose 5 independent numbers (regardless of the actual size of the set). There is only 1 possibility that the first number you select is less than the second less than the third, etc. for that distinct set of 5 that you selected.



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Re: M2706
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31 Aug 2015, 10:56
Thank you for the read. With a subset of N numbers it is clear to me that there are N! ways to arrange these numbers but that only 1 of the arrangements is ascending. But I had some difficulty understanding why answer B is correct. What I did not understand is why is does not matter how large Set A is. However, after plugging in some numbers I managed to convince myself. Two examples:
For k=3 assume A is the set 1, 2 and 3 and than n=2. So we select 2 numbers from a group of 3. Total ways to select 2 numbers onebyone when the order matters is 3x2=6. Ways to arrange two numbers in ascending order is 3: 1,2 2,3 1,3 Probability numbers selected in ascending order is 3/6=1/2
Now the same for k=4 where A is 1, 2, 3 and 4. Again select 2 numbers. Total ways to select 2 numbers 4x3=12. Ways to arrange numbers in ascending order is 6: 1,2 1,3 1,4 2,3 2,4 3,4 Probability numbers selected in ascending order is 6/12=1/2
So even though the set A is larger the probability is the same for the same n. Could someone confirm if my understanding and calculation is correct?



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Re M2706
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02 Sep 2015, 08:32
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Not very clear I'm afraid.Please elaborate on how B is sufficient here.



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What a beautiful question! The question is not about counting how many subsets A has, but about counting how many selections of n elements are made in ascending order. Once the selection is made the elements can be ordered in n! ways, the elements will be in ascending order in only one situation. So statement (1) is out of scope and only (2) can give the answer, by providing the value of n. Letter B.



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Re: M2706
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12 Oct 2015, 18:27
Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set onebyone, 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. i.e, \(k=12\) \(n\leq{k}\) so n can be any number ranging from 1 to 12. hence the probability will vary over \(1, \frac{1}{2}, \frac{1}{3}...........\frac{1}{12}\) (2) \(n=5\) \(k\geq{5}\) thus k can be any value from \([6,∞)\) indicating so many distinct numbers. But whatever be the no of items there, we are picking only 5 out of them these 5 can have \(5!\) combinations. out of these combinations, we have only way to order them in ascending order Therefore the probability that numbers will be selected in ascending order is \(\frac{1}{5!}\) So B is the answer.
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Re M2706
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07 Nov 2015, 04:07
I think this is a highquality question and I don't agree with the explanation. Doesn't statement 1 mean set A consists of 12 even consecutive integers only? as in n(A) = 12. In that case this statement would also be sufficient. Please elaborate.



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Re: M2706
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08 Nov 2015, 18:10
Bunuel wrote: Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set onebyone, 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) \(n=5\). Hi Gmatclub! I'm having a massively hard time understanding this solution haha. Went through the discussions here and still can't get it =( Any help would be appreciated! Point #1: is this saying that the probability of selecting n numbers from k numbers is always 1? For example, probability of selecting 5 numbers from a set of 10 number would be (10 choose 5)/(10 choose 5), so always 1? Point #2: we can select the n numbers in n! ways (i.e., if n =3, the number of ways to arrange n is 3*2*1 or 3!). Out of the n! ways, one 1 ways is ascending order, so the probability is 1/n! Thank you!!
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happyface101 wrote: Point #1: is this saying that the probability of selecting n numbers from k numbers is always 1? For example, probability of selecting 5 numbers from a set of 10 number would be (10 choose 5)/(10 choose 5), so always 1?
Well, this probability is centainly not 1. You can choose 0 numbers, 1 numbers, 2 numbers,..., 10 numbers. So the probablity you are saying would be equal to (10 choose 5)/(10 choose 0+10 choose 1 + 10 choose 2 + .... + 10 chosse 10). However, I don't see how can this be helpful in this problem... The problem already stated that n numbers were selected form the set and the problem is basically asking in how many ways this n numbers are selected in ascending order. happyface101 wrote: Point #2: we can select the n numbers in n! ways (i.e., if n =3, the number of ways to arrange n is 3*2*1 or 3!). Out of the n! ways, one 1 ways is ascending order, so the probability is 1/n!
Perfect, that's way (2) alone is correct.



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Re: M2706
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09 Nov 2015, 05:45
sashankmv wrote: I think this is a highquality question and I don't agree with the explanation. Doesn't statement 1 mean set A consists of 12 even consecutive integers only? as in n(A) = 12. In that case this statement would also be sufficient. Please elaborate. I think you are confusing k with n. Statement (1) says that k=12. How can this be helpful in answering how many ways the n numbers selected will be in ascending order? (1) is completly out of scope.



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Re M2706
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01 Feb 2016, 05:11
I think this is a highquality question and I don't agree with the explanation. For this question: I believe one needs to know both the size (k) of the Set A and the size (n) of the subset to answer this question. The probability of selecting 5 ascending numbers of a set of k=5 numbers is not the same as of the set where k=100. Therefore, I suggest the answer be C) "BOTH STATEMENTS"



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Re M2706
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16 Apr 2016, 08:47
I think this is a highquality question and I agree with explanation.



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Re M2706
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11 May 2016, 23:37
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. I cannot understand the answer explanation of this one. i went to the forum and it is confusing there as well. In the egmat class the formula for probability is clearly stated as x/n where x is the number of ways in which the event is possible and n is the total possible outcomes. in this case, n would be no. of ways of selecting 5 integers out of the total number of integers multiplied with no. of ways by which they can be arranged. i.e 12C5 x 5P5 if this is true, then how does the original set i.e. K not matter? the questions states ' what is the probability that the numbers will be 'selected' in an ascending order' ... and not 'arranged' in an ascending order!







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