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# M27-06

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Manager
Joined: 02 Mar 2017
Posts: 82
GMAT 1: 700 Q51 V34

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13 Jun 2017, 02:14
1
1
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) .
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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!

Intern
Joined: 04 Jul 2017
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18 Apr 2018, 08:06
1
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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Joined: 07 Feb 2018
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05 May 2018, 23:36
the question asks to select n numbers ONE BY ONE
note it is not a selection of n numbers at once so just think
if it the set has k elements,each element has equal probability to get selected, now i can even agree that every subset that has n elements has an equal probability to get selected

my doubt is regarding the 2nd point in the solution, you have already supposed that 5 elements are selected and they can be arranged in 5! ways and 1 out of these ways will be in ascending order - AGREED but what about the probablity of selecting that subset

asked probabilty = no of subsets in ascending order/total no of subsets
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04 Jun 2018, 14:03
I think this is a poor-quality question and I agree with explanation. This question should say integers not numbers
Math Expert
Joined: 02 Sep 2009
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04 Jun 2018, 21:40
I think this is a poor-quality question and I agree with explanation. This question should say integers not numbers

Not true. What does saying "integers" change?
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Joined: 24 May 2017
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20 Jun 2018, 03:56
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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Joined: 18 Feb 2018
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Location: New Zealand
Concentration: Strategy, Technology
WE: Consulting (Computer Software)

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27 Jun 2018, 22:07
Quote:
A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n≤k, what is the probability that numbers will be selected in ascending order?

1) Set A consists 12 even consecutive numbers
2) n= 5

Let’s say K comprised of (a,b,c,d,e) and

Case 1: n =3
Possible selection in ascending order - (a,b,c), (b,c,d), (c,d,e)
This is dependent on number of elements in K.
Case 2: n =5
Only one possible selection in ascending order is (a,b,c,d,e).
@bunuel- please let help understand why this is not option C

Posted from my mobile device
Manager
Joined: 25 Jan 2018
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Location: United States (IL)
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03 Jul 2018, 06:13
Hello Bunuel -

The sample space of selecting n numbers from K (one by one) should be k*k*K (n times right).??
how is it n factorial. Could you explain please

Thanks,
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13 Jul 2018, 07:25
I think this is a high-quality question.
Intern
Joined: 25 Nov 2017
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Location: India
Schools: IIMA , IIMB, IIMC
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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.

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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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Joined: 03 Apr 2018
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12 Sep 2018, 09:47
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 13 Jun 2017
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13 Sep 2018, 03:28
Hi,

Please can anyone let me know if the following approach is correct?

Event : Out of k integers, n integers are selected and arranged in ascending order.
Prob = (Number of ways n integers are selected from K integers and arranged in ascending order) / (Number of ways n integers are selected out of K integers and arranged in all possible orders).

Total ways: (K C N) * N! (For Simplification k = 12 and n = 5 , then 12C5 * 5!)
Number of Ways Event occurs : (K C N) * 1 (Only one way to arrange in ascending order) (12C5*1)
Prob = (KCN)*1/ (KCN)N! = 1/n! 12C5*1 /12C5*5! = 1/5!

Regards,
Akshay
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14 Sep 2018, 08:29
Akshay_Naik wrote:
Hi,

Please can anyone let me know if the following approach is correct?

Event : Out of k integers, n integers are selected and arranged in ascending order.
Prob = (Number of ways n integers are selected from K integers and arranged in ascending order) / (Number of ways n integers are selected out of K integers and arranged in all possible orders).

Total ways: (K C N) * N! (For Simplification k = 12 and n = 5 , then 12C5 * 5!)
Number of Ways Event occurs : (K C N) * 1 (Only one way to arrange in ascending order) (12C5*1)
Prob = (KCN)*1/ (KCN)N! = 1/n! 12C5*1 /12C5*5! = 1/5!

Regards,
Akshay

Part of the trap is doing unnecessary calculations on combinatorics.
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Joined: 28 Apr 2018
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17 Sep 2018, 12:49
Given that n≤k:
(1) If k =12 even consecutive integers , n could be any even positive integer from 2 to 24 depending on your assumption.Let's say a subset of set k contains one element i.e n=1,we can have subsets as {2},{4}....{24} in which case it becomes difficult to determine ASCENSION,hence it is insufficient. But if n=5 we can have subsets as {2,4,6,8,10},{3,2,4,24,6}... now it is obvious that we can determine the probability that the numbers in a subset are selected in ascending order, this is sufficient,but overall insufficient because of the inconsistency.

(2) n=5. Let's say k consists of 12 distinct numbers, then we can have a subset of k as {1,2,3,4,5} which allows us to determine the probability,hence this option is sufficient so B is the right answer.
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Joined: 12 Jan 2017
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20 Sep 2018, 11:22
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 $$A$$.

This is a good explanation! also first 95% difficulty i've gotten right in a while! woohoo
Re: M27-06 &nbs [#permalink] 20 Sep 2018, 11:22

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# M27-06

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