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Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3 ….., 20. What is the probability that 2 appears at an earlier position that any other even number in the selected permutation?
- a)1/2
- b)1/10
- c)9!/20!
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Suppose we uniformly and randomly select a permutation from the 20! pe...
If we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3,...,20, there are a few questions we can answer:
1. What is the probability that the number 1 is in the first position?
The probability that the number 1 is in the first position is 1/20, since there are 20 possible numbers that can be in the first position, and only one of them is 1.
2. What is the probability that the numbers 1, 2, and 3 appear in consecutive positions?
The probability that the numbers 1, 2, and 3 appear in consecutive positions is (20-2)! / 20!, which simplifies to 18! / 20!.
3. What is the expected value of the position of the number 1?
The expected value of the position of the number 1 is (1/20) * 1 + (1/20) * 2 + (1/20) * 3 + ... + (1/20) * 20, which simplifies to (1/20) * (1 + 2 + 3 + ... + 20). The sum of the numbers from 1 to 20 is (20 * 21) / 2 = 210, so the expected value is (1/20) * 210 = 10.5.
4. What is the probability that the number 1 is in a position greater than 10?
The probability that the number 1 is in a position greater than 10 is (20-10)! / 20!, which simplifies to 10! / 20!.
These are just a few examples of questions that can be answered about the randomly selected permutation. The specific probabilities and expected values will depend on the specific permutation and the question being asked.
1. What is the probability that the number 1 is in the first position?
The probability that the number 1 is in the first position is 1/20, since there are 20 possible numbers that can be in the first position, and only one of them is 1.
2. What is the probability that the numbers 1, 2, and 3 appear in consecutive positions?
The probability that the numbers 1, 2, and 3 appear in consecutive positions is (20-2)! / 20!, which simplifies to 18! / 20!.
3. What is the expected value of the position of the number 1?
The expected value of the position of the number 1 is (1/20) * 1 + (1/20) * 2 + (1/20) * 3 + ... + (1/20) * 20, which simplifies to (1/20) * (1 + 2 + 3 + ... + 20). The sum of the numbers from 1 to 20 is (20 * 21) / 2 = 210, so the expected value is (1/20) * 210 = 10.5.
4. What is the probability that the number 1 is in a position greater than 10?
The probability that the number 1 is in a position greater than 10 is (20-10)! / 20!, which simplifies to 10! / 20!.
These are just a few examples of questions that can be answered about the randomly selected permutation. The specific probabilities and expected values will depend on the specific permutation and the question being asked.
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Community Answer
Suppose we uniformly and randomly select a permutation from the 20! pe...
Number of permutations with ‘2’ in the first position = 19!
Number of permutations with ‘2’ in the second position = 10 × 18!
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
Number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by
Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.
Number of permutations with ‘2’ in the second position = 10 × 18!
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways)
Number of permutations with ‘2’ in 3rd position = 10 × 9 × 17!
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by
Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.
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