Number of Distinct Cycles of Length 3 in S5: 20

To find the number of distinct cycles of length 3 in S5, we can use the concept of permutations. In the symmetric group S5, each element represents a permutation of the numbers 1 to 5. A cycle of length 3 represents a permutation where three elements are cyclically rearranged.

To calculate the number of distinct cycles of length 3, we can consider the first element in the cycle as fixed and then count the number of ways to arrange the remaining 4 elements in a cycle.

Let's break down the calculation:

1. Choose the first element: We have 5 choices for the first element.

2. Arrange the remaining 4 elements: The remaining 4 elements can be arranged in (4-1)! = 3! = 6 ways.

3. Divide by the number of repetitions: Since the cycle is cyclic, the same cycle can be represented in multiple ways depending on the starting point. For a cycle of length 3, there are 3 possible starting points. So, we divide by 3.

Therefore, the number of distinct cycles of length 3 in S5 can be calculated as:
5 choices for the first element × 6 ways to arrange the remaining elements ÷ 3 possible starting points = 10.

Hence, option A is correct.

Number of Distinct Cycles of Length 2 in S5: 10

Similarly, to find the number of distinct cycles of length 2 in S5, we can follow the same approach.

1. Choose the first element: We have 5 choices for the first element.

2. Arrange the remaining 3 elements: The remaining 3 elements can be arranged in (3-1)! = 2! = 2 ways.

3. Divide by the number of repetitions: Since the cycle is cyclic, the same cycle can be represented in multiple ways depending on the starting point. For a cycle of length 2, there are 2 possible starting points. So, we divide by 2.

Therefore, the number of distinct cycles of length 2 in S5 can be calculated as:
5 choices for the first element × 2 ways to arrange the remaining elements ÷ 2 possible starting points = 5.

Hence, option B is correct.

Number of Elements of Order 6 in S5: 20

The order of an element in a permutation group is the smallest positive integer k such that raising the element to the power of k results in the identity permutation.

To find the number of elements of order 6 in S5, we need to consider the cycles of length 6.

1. Choose the first element: We have 5 choices for the first element.

2. Arrange the remaining 4 elements: The remaining 4 elements can be arranged in (4-1)! = 3! = 6 ways.

3. Divide by the number of repetitions: Since the cycle is cyclic, the same cycle can be represented in multiple ways depending on the starting point. For a cycle of length 6, there are 6 possible starting points. So, we divide by 6.

Therefore, the number of elements of order 6 in S5 can be calculated as:
5 choices for the first element × 6 ways to arrange the remaining elements ÷ 6 possible starting points = 5.

Hence