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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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12 Answers

2 votes
2 votes

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 votes
1 votes

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

1 votes
1 votes
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.
1 votes
1 votes

First consider all odd numbers: select all odd numbers and arrange them

                                                      =>20P10 ways.

Now consider 2: Fix 2 in first place available after arranging all odd numbers

                            => 1 way.

Now consider even numbers other than 2: arrange all 9 of those even numbers 

                                                                       => 9! ways

Total ways of arranging numbers when 2 at an earlier position than any other even number: 20P2 * 1 * 9! ways

Total ways of arranging 20 numbers: 20! ways

=> Probability = $\frac{20P10 * 1 * 9!}{20!}$

                          = $\frac{20! * 9!}{10! * 20!}$

                          = $\frac{1}{10}$

Answer:

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