Answer:
(a) 6 permutations
(b) 6 permutations
(c) 24 permutations
Explanation:
Given the set {1, 2, 3, 4, 5, 6}
We need to check how many permutations exist such that:
(a) No even integer is in its natural position.
All we need to do is count the number of elements, n, between which we want to make permutations, and take factorial of n to know the number of permutations.
The even integers are 2, 4, and 6, and we want to ensure that they are in their natural positions. So, the permutations are between 3 elements. For n = 3, factorial of 3 is 3! = 3×2×1 = 6.
To back this up, see the permutations.
{1, 2, 3, 4, 5, 6}, {1, 2, 5, 4, 3, 6}, {3, 2, 5, 4, 1, 6}, {3, 2, 1, 4, 5, 6}, {5, 2, 1, 4, 3, 6}, and {5, 2, 3, 4, 1, 6}
Six permutations.
(b) Exactly three integers are in their natural position.
Exactly 3 integers leaves us with permutations between three numbers, so 3! = 6.
We have six permutations.
(c) 1 and 6 are in their natural positions
If only 1 and 6 are in there natural position, then we have permutations between 4 elements.
We have 4! Permutations.
4! = 4×3×2×1 = 24