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Solve the following question and mark the best possible option.
Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?
- a)24
- b)45
- c)43
- d)44
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Solve the following question and mark the best possible option.Kamal a...
This is a case of derangement. In combinatorial mathematics, a derangement is a permutation of the elements of a set such that none of the elements appear in their original position. The derangement formula is 5!(1/2! − 1/3! - 1/4! + 1/5! ) Using it here we get the answer as 44.
Hence, option 4.
Hence, option 4.
Alternate solution:
The total number of ways of arranging the 5 digits is 5! = 120. We are not interested in arrangements where 1 or more of the digits occupy the corresponding place, i.e., we are not interested in 5C, 3C2W, 2C3W or 1C4W. All 5 digits occupying the correct corresponding places, i.e., 5C, can happen in only 1 way. In case of 3C2W, the 3 correct cases can be chosen in 5C3 = 10 ways and the 2 wrong digits can be arranged in 1 way. So, there are 10 possibilities for 3C2W. In case of 2C3W, the 2 correct digits can be chosen in 5C2 = 10 ways and the 3 wrong digits can be arranged in 2 ways. So, there are 10 × 2 = 20 possibilities for 2C3W. In case of 1C4W, the correct digit can be chosen in 5 ways and the 4 wrong digits can be arranged in 9 ways. So, there are 5 × 9 = 45 possibilities for 1C4W. Thus, the number of ways in which none of the digits occupy their corresponding positions is 120 – 1 – 10 – 20 – 45 = 44.
The total number of ways of arranging the 5 digits is 5! = 120. We are not interested in arrangements where 1 or more of the digits occupy the corresponding place, i.e., we are not interested in 5C, 3C2W, 2C3W or 1C4W. All 5 digits occupying the correct corresponding places, i.e., 5C, can happen in only 1 way. In case of 3C2W, the 3 correct cases can be chosen in 5C3 = 10 ways and the 2 wrong digits can be arranged in 1 way. So, there are 10 possibilities for 3C2W. In case of 2C3W, the 2 correct digits can be chosen in 5C2 = 10 ways and the 3 wrong digits can be arranged in 2 ways. So, there are 10 × 2 = 20 possibilities for 2C3W. In case of 1C4W, the correct digit can be chosen in 5 ways and the 4 wrong digits can be arranged in 9 ways. So, there are 5 × 9 = 45 possibilities for 1C4W. Thus, the number of ways in which none of the digits occupy their corresponding positions is 120 – 1 – 10 – 20 – 45 = 44.
Most Upvoted Answer
Solve the following question and mark the best possible option.Kamal a...
Problem Analysis:
We need to find the number of ways in which the numbers 1, 2, 3, 4, and 5 can be rearranged without repetition such that the number n is never in the nth position. In other words, we need to find the number of derangements of the numbers 1, 2, 3, 4, and 5.
Solution:
To solve this problem, we can use the concept of derangements. A derangement is a permutation of the elements of a set, such that no element appears in its original position.
Step 1: Calculate the factorial of 5 (the total number of elements).
5! = 5 x 4 x 3 x 2 x 1 = 120
Step 2: Calculate the sum of alternating signs of the factorials.
(-1)^0 x 5! + (-1)^1 x 4! + (-1)^2 x 3! + (-1)^3 x 2! + (-1)^4 x 1! = 120 - 24 + 6 - 2 + 1 = 101
Step 3: Calculate the number of derangements.
The number of derangements is given by the formula:
D(n) = n! x (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)
For n = 5, the number of derangements is:
D(5) = 5! x (1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 -
We need to find the number of ways in which the numbers 1, 2, 3, 4, and 5 can be rearranged without repetition such that the number n is never in the nth position. In other words, we need to find the number of derangements of the numbers 1, 2, 3, 4, and 5.
Solution:
To solve this problem, we can use the concept of derangements. A derangement is a permutation of the elements of a set, such that no element appears in its original position.
Step 1: Calculate the factorial of 5 (the total number of elements).
5! = 5 x 4 x 3 x 2 x 1 = 120
Step 2: Calculate the sum of alternating signs of the factorials.
(-1)^0 x 5! + (-1)^1 x 4! + (-1)^2 x 3! + (-1)^3 x 2! + (-1)^4 x 1! = 120 - 24 + 6 - 2 + 1 = 101
Step 3: Calculate the number of derangements.
The number of derangements is given by the formula:
D(n) = n! x (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)
For n = 5, the number of derangements is:
D(5) = 5! x (1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
= 120 x (1 - 1 + 1/2 -
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Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer?
Question Description
Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer?.
Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer?.
Solutions for Solve the following question and mark the best possible option.Kamal asked Rahim to find the number of ways in which all the numbers 1, 2, 3, 4 and 5 can be rearranged without repetition such that the number n is never in the nth position. For example, 25413 and 51234 are acceptable, but 12435 is not. How many such rearrangements are possible?a)24b)45c)43d)44Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for CAT.
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