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M27-06

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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) .
.
.
.
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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!

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 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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New post 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
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New post 04 Jun 2018, 21:40
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New post 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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New post 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

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New post 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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New post 13 Jul 2018, 07:25
I think this is a high-quality question.
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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!.
Re: M27-06 &nbs [#permalink] 31 Jul 2018, 03:54

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