How many permutations of {1, 2, 3, 4, 5, 6} exist such that: (a) No even integer is in its natural position. (b) Exactly three integers are in their natural position. (c) 1 and 6 are in their natural positions.

(a) 6 permutations

(b) 6 permutations

(c) 24 permutations

Step-by-step 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

2

step-by-step explanation:

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )