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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

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.
edited

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

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

https://youtu.be/GzFZvTBztXw

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: $B$
by

good explanation must see this video to understand the problem
Thanks sir

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

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

by

Nice explanation ..
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

@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 .

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

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

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.

by
 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