M27-06

Official Solution:


It is important to note two things:

1. The probability of selecting any subset of \(n\) numbers from the set is the same. No particular subset of \(n\) numbers is favored over any other subset of \(n\) numbers, so they all have an equal probability of being selected.

2. Any subset of \(n\) distinct numbers can be selected in \(n!\) ways, and only one out of these \(n!\) possible ways will be in ascending order. For example, subset of {1, 2, 3} can be selected in 6 ways: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) or (3, 2, 1). As you can see, out of these \(3!\) possible ways, only one is in ascending order, (1, 2, 3). Therefore, the probability of selecting a subset in ascending order is \(\frac{1}{n!}\).

Thus, to answer the given question, we only need to know the size of the subset (\(n\)) that we are selecting from the set \(A\). Therefore, the firs statement is not sufficient, while the second one is.


Answer: B
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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.

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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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.

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.

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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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

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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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

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.

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!.

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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ALTERNATE EXPLANATION:


\(Probability = \frac{Fav\:Outcomes}{Total\:Outcomes}\)

1) Total Number of outcomes

Total Number of outcomes = Number of ways 5 numbers can be picked & arranged from 12 numbers = \(^{12}P_5\)

2) Total Number of Fav outcomes....This is where it gets tricky

Each subset of 5 numbers consists of only one arrangement in which the numbers are there in ascending order...

For example:
Consider the subset {1,2,3,4,5}
This can be arranged in many ways {1,2,3,4,5} , {1,3,4,2,5} & so on..
But note that only one arrangement in the set {1,2,3,4,5} is in ascending order out of all arrangements.

Similarly, the same happens with subset {1,2,3,4,6}.
Only one subset will be in ascending order for each subset.

Thus, we need to find how many subsets of 5 are possible. That is a COMBINATIONS sequence.

Total Number of Fav outcomes = Number of ways of choosing 5 out of 12 numbers = \(^{12}C_5\)

Thus, \(Probability =\frac{^{12}C_5}{^{12}P_5}\)

We can write in a generalized way, where n is the number of elements and r is the number of selections.

Thus,Probability =\(\frac{ ^{n}C_r}{^{n}P_r}\) = \(\frac{\frac{n!}{(n-r)!r!}}{\frac{n!}{(n-r)!}}\) =\( \frac{n!}{(n-r)!r!}*\frac{(n-r)!}{n!}\) = \(\frac{1}{r!}\)

Thus, only r matters. r is the number of elements that is selected. This, is denoted as n in the question.
And, the value of n is given in (B).

Thus, (B) is sufficient.



+1 Kudos if you like my explanation. I broke my head for this.

Bunuel

we are looking for the probability that the 5 numbers picked ( picked one by one) are in ascending order,

If we know that n=5, we can have at the 1st pick x1, at the 2nd pick x2 at the 3rd pick x3, at the 4th pick x5 we'll then have a selection of (x1,x2,x3,x4,x5) this is a case of ascending order

If actually we pick at first x1, at 2nd x2 at 3rd x3 at 4th x4 but at 5th x6, then we would have picked a selection ( x1,x2,x3,x4,x6) which is a selection also in ascending order,

So we have more than one case of a selection in ascending order !!!

Can you explain

KarishmaB i have the same question as the post above (highlighted) . can you pls explain