GATE CSE 2007 | Question: 24

Question

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?

• A. $\left(\dfrac{1}{2} \right)$
• B. $\left(\dfrac{1}{10}\right)$
• C. $\left(\dfrac{9!}{20!}\right)$
• D. None of these

Comments

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.

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Great explanation by Pankaj Porwal Sir

Total number of places = 20 (to put numbers 1 to 20)

1. Select 10 places out of 20 to place odd numbers (#10 odd nums) = 20C10 ways. 

Arrange all odd numbers here

= 20C10 * 10!

2. Places left = 10 ( we needa arrange 10 even numbers here)

& 2 must be before all even numbers.

Therefore, only one(1) place for 2.

Now arrange left 9 odd numbers in leftover 9 places i.e. 9! ways

Answer = 

(20C10 * 10! * 1 * 9!)/(TotalWays ie 20!) = 1/10

https://youtu.be/GzFZvTBztXw

Answer 1

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: $B$

Comments

Thanks sir

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@Arjun sir, Can you please refer some more questions related to the concept that you explained here?

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Made Easy is wrong… the expression can be reduced to gat 1/10 as follows:

Answer 2

Total possible permutations $= 20!$

For $2$ coming before any other even number, first we have to fix the positions of $10$ even numbers which can be done in ${}^{20}C_{10}$ ways. Now even numbers other than $2$ can be permuted in $9!$ ways and $10$ odd numbers (they have only $10$ places left) can be permuted in $10!$ ways.

So, number of ways in which $2$ comes before any other even number $={}^{20}C_{10} *10! 9!$

Required probability $=\dfrac{{}^{20}C_{10} .10!. 9!}{20!} =1/10.$

Answer is (B)

Comments

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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[email protected] now.

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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 .

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Just some notes i made, if it's helpful. 

Answer 3

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.


Please check ....

Answer 4

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