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Suppose we uniformly and randomly select a permutation from the $20 !$ permutations of $1, 2, 3\ldots ,20.$ What is the probability that $2$ appears at an earlier position than any other even number in the selected permutation?

1. $\left(\dfrac{1}{2} \right)$
2. $\left(\dfrac{1}{10}\right)$
3. $\left(\dfrac{9!}{20!}\right)$
4. None of these

edited | 5.5k views
+3
Hi Guys,

Just small information some people have posted some MadeEasy solution. I would like to say their approach is correct but answer is wrong. If you will solve their equation then also you will get option (B) as answer.

(1/20)+(1/38)+(1/76)+(2/(19*17))+(7/(19*17*8))+(7/(20*19*17))+(1/(19*17*8))+(1/(26*19*17))+(1/(26*19*17*4))+(1/(26*22*19*17))+(1/(26*22*19*17*10)) = 1/10

But in exam i think this aproach will take more time. So follow the approach as suggested by @Arjun sir.

There are $10$ even numbers $(2,4\ldots 20)$ possible as the one in the earliest position and all of these are equally likely. So, the probability of $2$ becoming the earliest is simply $\dfrac{1}{10}$.

Correct Answer: $B$
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@Arjun can you please point out why the following solution is wrong, thanks.

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@arjun sir

2 appears at an earlier position than any other even numbe?
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Their approach is correct but specified answer is wrong.
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@Arjun Sir, I didn't quite get the question. What is meant when it says

'2 appears at an earlier position than any other even number in the selected permutation?' What is the position here? Is it the position in the value of a permutation of any no between 1 to 20? Please clarify.
+3

we have numbers from 1 - 20
Question says that their permutation (arrangement) is 20! (that is obvious)
i.e like 1,2,3 .......20
2,3,4 ......1,20
Like this we have 20! arrangements

Now,

'2 appears at an earlier position than any other even number in the selected permutation?'
Means
That the no. 2 should appear before  all other even no. in that particular arrangement you have selected.

+1

@Arjun  Sir, I am unable to get the solution,

In rosen,symmetry of number has mentioned

2 being earlier in the set of even numbers is 2 is placed in first position , rest all can be in 9!   so it will be 1*9!/10! =1/10

Is this way correct, [ in any no a,b   a->b is same as b->a ]

+17

2 ${\color{Red} \star }$4 ${\color{Red} \star }$6${\color{Red} \star }$ 8${\color{Red} \star }$ 10 ${\color{Red} \star }$12 ${\color{Red} \star }$14 ${\color{Red} \star }$16 ${\color{Red} \star }$18 ${\color{Red} \star }$20${\color{Red} \star }$

Since we want 2 to appear before any other even number in the permutation, the permutation of odd numbers is insignificant and we can consider the above case where 2 can occupy total of 10 positions considering some random permutation of even numbers.

Now since only 10 positions are there and out of which only 1 position we have in which 2 comes before any other even number so probability=1/10.

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a great approach arjun sir
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What is the probability that 2 appears at an earlier position than any other  number in the selected permutation? if this is a question then according to approach specified by arjun sir answer will be (1/20),,.correct me if i am wrong
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Since it is a uniform distribution of permutations/vectors, can we see it this way? - we can divide the set of permutations into 10 equally sized groups (think: dividing the interval of the uniform distribution into 10 equal length subintervals). This 10-group subdivision is according to whichever even number gets to be in the first position, out of the 10 even numbers available. There is only one favorable group/outcome in this sample space, the one with 2 in the lead, so the probability becomes $1/10$.

The reason I even thought of this is because I had noticed a pattern while doing uniform distribution questions. For example there is https://gateoverflow.in/1694 where @Arjun sir divides the uniform interval of 60 minutes into subintervals of 10 minutes each to solve the question.

I don't know if this is correct. This doc is all I could find: http://www.math.drexel.edu/~tolya/continuous%20uniform.pdf and it says that -:

A random variable X taking values in the interval [a, b] is uniformly distributed if in any two equal subintervals of [a, b] X occurs with the same probability.

That's the closest I could get to "subdivision" in the definition of uniform distribution.

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Right!

Total permutation= 20!

2 come before any other number= first we have to fix position for 10 even number so we can do it in 20C10 ways. Now other than 2 can permuted in 9! ways and 10 odd num can be permuted in 10! ways

so Ways in which 2 comes before any other even number is-  20C10 *10! 9!

So probability= 20C10 *10! 9!/20! =1/10.

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Just small correction in place of  20C10 *10!89! it should be  20C10 *10!*9! = 20! / 10

So the prob = (20! / 10) / 20! =1/10

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When you are separately doing 10! and 9!, isn't it indicating that the even nos. are separately getting rearranged among themselves and odd nos. are getting rearranged among themselves only. For eg. if 1,2,3,4 is the given set, then it is like (1,2,3,4), (2,1,3,4), (2,1,4,3), (1,2,4,3). It does not include the cases like (2,3,1,4) or (1,4,2,3) and so on, that is the intermixing of odd and even numbers.
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Nice explanation ..
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1, 2, 3, 4, …….20
Total no. of possible even number = 10
Here we are not considering odd number.
The probability that 2 appears at an earlier position than any other even number is =1/10
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@Krithiga2101

but after placing 10 even number we have total 11 positions to place 10 odd . then why don't you first choose 10 position from 11 available position then . arrange them in 10! ways .

 The odd numbers do not matter here. The probability 2 comes before the other 9 evens is (# of ways to pick 2)(# of ways to pick remaining evens)/(# of ways to order 10 evens) 1*9!/10!=1/10
Answer is 1/10 .
(read above made easy solution image attached and read Amitabh tiwari comment )

Now My solution-

Made easy solution is correct , if you solve this ( a big calculation) you'll get 1/10 as answer.
So, here calculation is so big so we will solve a small example and try to find out pattern.

Example 1- if i have four numbers 1,2,3,4.  then total permutation is 4! .

Now, same condition is here  " What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?"

now ,  we will solve using method
Number of permutations with 2 in first position    -->   2  _  _  _  =   3!=  6

Number of permutations with 2 in second position    -->   _   2   _   _ =  2 *2!=  4

Number of permutations with 2 in third position    -->    _  _  2  _  =  2*1*1= 2

so probability is =    6+4+2  / 4!   =  12 /24 = 1/2

So Answer is 1/2 .

Second method as Arjun Sir explained -

There are 2 even numbers (2,4) possible as the one in the earliest position and all of these are equally likely. So, the

probability of 2 becoming the earliest is simply 1/2.

Example 2- if i have Six numbers 1,2,3,4,5,6.  then total permutation is 6! .

Now, same condition is here  " What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?"

now ,  we will solve using method
Number of permutations with 2 in first position    -->   2  _  _  _  _  _ =   5!

Number of permutations with 2 in second position    -->   _   2   _   _  _  _ =  3 *4!

Number of permutations with 2 in third position    -->    _  _  2  _   _   _=  3*2*3!

Number of permutations with 2 in Fourth position    -->    _  _  _  2   _   _= 3*2*1*2!

so probability is =  5! + 3*4! + 3*2 *3! + 3*2*1*2! / 6!  =  40 / 120 = 1/3

So Answer is 1/3 .

Second method as Arjun Sir explained -

There are 3 even numbers (2,4,6) possible as the one in the earliest position and all of these are equally likely. So, the

probability of 2 becoming the earliest is simply 1/3.

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Here order of odd numbers doesn't matter, so we focus only on 10 even numbers as if we have to arrange only 10 even numbers . In general, digit 2 can be placed at any of the 10 places available, but according to question, 2 can be placed at only 1st because it has to appear before every other even number. So out of 10 choices, we have only 1 favourable choice, so probability is 1/10. So option (B) is correct.

+1 vote

This solution is given in Made easy book. please verify :(

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madeeasy solution can be verified only by madeeasy.
hifi solution
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lol.
+4
This solution is absolutely correct!. Even this will lead to 1/10.

try with 1,2,3,4 using this technique,  you will get it
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Their approach is correct but specified answer is wrong.
+1 vote
All even numbers have the same probability of being first (The odd numbers do not matter here). And there are 10 of them. The probability 2 comes before the other 9 evens is = 1/10.
So, option (B) is correct.
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Let's consider some cases where 2 is before every other even number:

Case 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Case 2: 1 3 5 7 9 11 13 15 17 19 4 6 8 10 12 14 16 18 20

Case 3: 2 1 3 5 7 9 11 13 15 17 19 4 6 8 10 12 14 16 18 20

As you can see, the positioning of odd numbers don't really matter as we are concerned with just the positioning of the subset of even numbers

Case 4 (just even numbers): 2 4 6 8 10 12 14 16 18 20

2 has to be in the first position. Any of these even numbers could be considered in the first position therefore the probability of 2 occurring before every other even number is 1/10.

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