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M27-06

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New post 16 Sep 2014, 01:27
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New post 16 Sep 2014, 01:27
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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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New post 13 Nov 2014, 07:27
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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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New post 14 Nov 2014, 02:51
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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 5-number 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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New post 29 Jul 2015, 18:03
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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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New post 10 Oct 2015, 10:56
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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: M27-06  [#permalink]

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New post 12 Oct 2015, 18:27
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Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set one-by-one, 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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New post 07 Nov 2015, 04:07
I think this is a high-quality 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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New post 08 Nov 2015, 18:10
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Bunuel wrote:
Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set one-by-one, 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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New post 09 Nov 2015, 05:39
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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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New post 09 Nov 2015, 05:45
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sashankmv wrote:
I think this is a high-quality 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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New post 11 May 2016, 23:37
I think this is a poor-quality 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 e-gmat 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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New post 12 May 2016, 00:04
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sheetaldodani wrote:
I think this is a poor-quality 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 e-gmat 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!


hi,
it is a good question..

when you have five items, they can be arranged in say x ways by whatever way you find it, BUT ONLY 1 out of entire ways is in ascending order...

say lets go by your way..
total ways possible = kC5*5P5..
now how many ways will the 5 in ascending order, 1 way of each 5 selected = kC5..
prob = \(\frac{kC5}{kC5*5P5} = 1/5P5\)..
so you see it doesn't matter WHAT k is, it depend on the number we are choosing out of k...
Now even if you have 100 items and you are choosing 5 out of them, these 5 will be arranged in 5! ways and 1 out of these 5! will be in ascending order
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New post 24 Apr 2017, 19:32
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I got this question wrong, but I guess I get the idea now.
Here is what I understood (Pardon the formatting) :
Numerator = Favourable outcomes. Selection and Arrangement, because order is important.
Number of ways to select n things out of k things = KCn
Number of ways to arrange the selection in ascending order = 1 way
Numerator = Selection and Arrangement = KCn * 1
Denominator = Total Outcomes. Selection and arrangement
Selection = Number of ways to select n things out of K things = KCn
Arrangement = Number of ways to arrange n things = n!
Probability = Favourable outcomes / Total Outcomes.
Probability = (KCn *1) / (KCn * n!) = 1/n!
Hence Option B
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New post 12 Jun 2017, 12:06
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I have a doubt on this one (and many others as well), but I guess if we are to select a subset of n numbers from a set with k distinct numbers we should have k!/(k-n)! ways to select and not just n!. If we are to choose 3 elements from a set of 5 elements we have 5*4*3.

In this case we would need both, k and n values, no?!

Thank you for further explanations! (Y)
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New post 12 Jun 2017, 13:20
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leandrodijon wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I have a doubt on this one (and many others as well), but I guess if we are to select a subset of n numbers from a set with k distinct numbers we should have k!/(k-n)! ways to select and not just n!. If we are to choose 3 elements from a set of 5 elements we have 5*4*3.

In this case we would need both, k and n values, no?!

Thank you for further explanations! (Y)


First of all, this is a hard question so you should be comfortable with combinations and probability to solve it.

Next, if we are choosing 5 numbers, the probability of choosing them in ascending order is 1/5!.

The number of permutations of 5 different number, the number of ways we can choose 5 different number, say 1, 2, 3, 4 and 5 is 5! = 120:
{1, 2, 3, 4, 5}
{1, 2, 3, 5, 4}
{1, 2, 4, 3, 5}
{1, 2, 4, 5, 3}
{1, 2, 5, 3, 4}
{1, 2, 5, 4, 3}
{1, 3, 2, 4, 5}
{1, 3, 2, 5, 4}
{1, 3, 4, 2, 5}
{1, 3, 4, 5, 2}
{1, 3, 5, 2, 4}
{1, 3, 5, 4, 2}
{1, 4, 2, 3, 5}
{1, 4, 2, 5, 3}
{1, 4, 3, 2, 5}
{1, 4, 3, 5, 2}
{1, 4, 5, 2, 3}
{1, 4, 5, 3, 2}
{1, 5, 2, 3, 4}
{1, 5, 2, 4, 3}
{1, 5, 3, 2, 4}
{1, 5, 3, 4, 2}
{1, 5, 4, 2, 3}
{1, 5, 4, 3, 2}
{2, 1, 3, 4, 5}
{2, 1, 3, 5, 4}
{2, 1, 4, 3, 5}
{2, 1, 4, 5, 3}
{2, 1, 5, 3, 4}
{2, 1, 5, 4, 3}
{2, 3, 1, 4, 5}
{2, 3, 1, 5, 4}
{2, 3, 4, 1, 5}
{2, 3, 4, 5, 1}
{2, 3, 5, 1, 4}
{2, 3, 5, 4, 1}
{2, 4, 1, 3, 5}
{2, 4, 1, 5, 3}
{2, 4, 3, 1, 5}
{2, 4, 3, 5, 1}
{2, 4, 5, 1, 3}
{2, 4, 5, 3, 1}
{2, 5, 1, 3, 4}
{2, 5, 1, 4, 3}
{2, 5, 3, 1, 4}
{2, 5, 3, 4, 1}
{2, 5, 4, 1, 3}
{2, 5, 4, 3, 1}
{3, 1, 2, 4, 5}
{3, 1, 2, 5, 4}
{3, 1, 4, 2, 5}
{3, 1, 4, 5, 2}
{3, 1, 5, 2, 4}
{3, 1, 5, 4, 2}
{3, 2, 1, 4, 5}
{3, 2, 1, 5, 4}
{3, 2, 4, 1, 5}
{3, 2, 4, 5, 1}
{3, 2, 5, 1, 4}
{3, 2, 5, 4, 1}
{3, 4, 1, 2, 5}
{3, 4, 1, 5, 2}
{3, 4, 2, 1, 5}
{3, 4, 2, 5, 1}
{3, 4, 5, 1, 2}
{3, 4, 5, 2, 1}
{3, 5, 1, 2, 4}
{3, 5, 1, 4, 2}
{3, 5, 2, 1, 4}
{3, 5, 2, 4, 1}
{3, 5, 4, 1, 2}
{3, 5, 4, 2, 1}
{4, 1, 2, 3, 5}
{4, 1, 2, 5, 3}
{4, 1, 3, 2, 5}
{4, 1, 3, 5, 2}
{4, 1, 5, 2, 3}
{4, 1, 5, 3, 2}
{4, 2, 1, 3, 5}
{4, 2, 1, 5, 3}
{4, 2, 3, 1, 5}
{4, 2, 3, 5, 1}
{4, 2, 5, 1, 3}
{4, 2, 5, 3, 1}
{4, 3, 1, 2, 5}
{4, 3, 1, 5, 2}
{4, 3, 2, 1, 5}
{4, 3, 2, 5, 1}
{4, 3, 5, 1, 2}
{4, 3, 5, 2, 1}
{4, 5, 1, 2, 3}
{4, 5, 1, 3, 2}
{4, 5, 2, 1, 3}
{4, 5, 2, 3, 1}
{4, 5, 3, 1, 2}
{4, 5, 3, 2, 1}
{5, 1, 2, 3, 4}
{5, 1, 2, 4, 3}
{5, 1, 3, 2, 4}
{5, 1, 3, 4, 2}
{5, 1, 4, 2, 3}
{5, 1, 4, 3, 2}
{5, 2, 1, 3, 4}
{5, 2, 1, 4, 3}
{5, 2, 3, 1, 4}
{5, 2, 3, 4, 1}
{5, 2, 4, 1, 3}
{5, 2, 4, 3, 1}
{5, 3, 1, 2, 4}
{5, 3, 1, 4, 2}
{5, 3, 2, 1, 4}
{5, 3, 2, 4, 1}
{5, 3, 4, 1, 2}
{5, 3, 4, 2, 1}
{5, 4, 1, 2, 3}
{5, 4, 1, 3, 2}
{5, 4, 2, 1, 3}
{5, 4, 2, 3, 1}
{5, 4, 3, 1, 2}
{5, 4, 3, 2, 1}

(total: 120)

Only one of the total of 120 arrangements is in ascending order, namely the first one: {1, 2, 3, 4, 5}. So, the probability is 1/120.
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New post 13 Jun 2017, 02:14
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Number of sets possible will depend both on K and N. So if K=6 and n= 5, then (6C5 * 5!) will be the number of subset possible. In general it is = KCN ( number of ways of selecting N terms from K terms) * N! ( permute the numbers selected).

Now number of cases in which the numbers are in ascending order is equal to 1 * KCN ( number of ways of selecting N terms from K terms). Why? Because for every N numbers selected from K terms there will be 1 subset which will be in ascending order.

eg. K( 6 terms)= {1,2,3,4,5,6}
N(5 terms) = {1,2,3,4,5} ( this general set has 5! sets 2,3,1,4,5 / 2,1,3,4,5 etc) .
N ( 5 terms) = {1,3,4,5,6}( this general set has 5! sets 6,3,1,4,5 / 6,1,3,4,5 etc) .
.
.
.
.
or N (5 terms) = {2,3,4,5,6} ( this general set has 5! sets 2,3,6,4,5/ 2,6,3,4,5 etc)

Hence in every general set there will be 1 set which is in ascending order. And number of general set is KCN.

Hence probability= KCN / ( KCN * N!) = 1/N!

Hence answer is B
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New post 18 Apr 2018, 08:06
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It can be in ascending order only after arrangement..

The way to select the n no out of k will be same kCn .the Q says that the selected no. are in ascending or not..so out of the selection there is only one way when the no. selected is in ascending order out of n! ways of arrangement.
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New post 17 Jul 2018, 11:29
Excellent question.
1) It gives value of K=12, then what is the value of n ? Value of n=1,2,3,.......12.
Let n=3, it means there are 3 numbers in this selection.
How may ways this selection (of 3 numbers) can be arranged ?
=> It is in 3! ways
How many ways these arrangements can be in an ascending order?
=> Only one way.
Probability is=1/3!
Let n=5 then probability =1/5! , n=6, probability=1/6! and so on... till n=12
Hence for different values of n, we will get different probabilities.
Hence statement 1 is not sufficient.


2) It says n=5, (So we are not worried about K, since we need to select 'n' numbers from set of K numbers )
Probability= 1/5!
( See in the statement 1, the value of n is not fixed.It can be anything from 1 to 12, Here in statement 2 , n=5)

It is sufficient.

Hence Answer is B.
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New post 31 Jul 2018, 03:54
Indeed a great question. If one gets the logic right, it comes to instantaneously.

As per B, it shall be thought of as if:
we have taken out 5 numbers from some Set. If we keep these 5 numbers (arrange them in different ways), what is the probability of them being in ascending order. And since all numbers are different there is only ONE way that they will be in ascending order. 5 numbers can be arranged in 5! ways. And hence 1/5!.
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