Question

If A= {1,2,3,-2n} then find the number of derangements of the elements of A such that the first n elements of each of the derangements are the last n elements of A .

Answer 1

Step-by-step explanation:

Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of {0, 1, 2, 3} is {2, 3, 1, 0}.

Given a number n, find the total number of Derangements of a set of n elements.

Examples :

Input: n = 2

Output: 1

For two elements say {0, 1}, there is only one

possible derangement {1, 0}

Input: n = 3

Output: 2

For three elements say {0, 1, 2}, there are two

possible derangements {2, 0, 1} and {1, 2, 0}

Input: n = 4

Output: 9

For four elements say {0, 1, 2, 3}, there are 9

possible derangements {1, 0, 3, 2} {1, 2, 3, 0}

{1, 3, 0, 2}, {2, 3, 0, 1}, {2, 0, 3, 1}, {2, 3,

1, 0}, {3, 0, 1, 2}, {3, 2, 0, 1} and {3, 2, 1, 0}

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Let countDer(n) be count of derangements for n elements. Below is the recursive relation to it.

countDer(n) = (n - 1) * [countDer(n - 1)