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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
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Total possible permutation = $20! $

there are 10 even numbers, if we stop permutations of all even numbers, and sort them in ascending order, we have $\frac{20!}{10!}$ permutations, now keeping 2 still permute rest 9 even numbers, no of permuation in which 2 is first amongst all even numbers to occur is $\frac{20!}{10!} \cdot 9!$

Required probability = $\frac{\frac{20!}{10!} \cdot 9!}{20!} = \frac{9!}{10\cdot 9!} = \frac{1}{10}$
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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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0 votes
The logic Behind solving this question is quite simple. the even number 2,4,6,8,10,12,14,16,18,20 are 10.

and the possible positions we can place “2” relative to  any even number(any permutation of the rest of the 9 numbers). is 10. Now just think about it in a broad sense that if we put the probability of fiding 2 before any even number occurs is X. Then the probability of “2” occuring before the  “8 even number and after first even number” will be the X too.

Similarly, 2-9even, 1even-2-8even, 2even-2-7even, 3even-2-6even, 4even-2-5even………... all will have probability X.

total number of such arrangement are 10, so 10X and the one that favours our condition is only one so X.

so the probability would be= X/10X

=1/10

Answer is B
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Sample space: Selecting 10 places for even numbers and they can permute in 10! ways = $\binom{20}{10}*10!$

Event: Selecting 10 places for even numbers BUT this time they can permute in only 9! ways because number 2 is fixed = $\binom{20}{10}*9!$

Probability = $\frac{\binom{20}{10}*9!}{\binom{20}{10}*10!}$ = $\frac{1}{10}$

Answer:

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