The problem asks for the number of elements of order 3 in the symmetric group $S_6$. The order of an element in $S_n$ is the least common multiple (LCM) of the lengths of the disjoint cycles in its cycle decomposition.
Identifying Cycle Structures for Order 3 Elements
An element in $S_6$ has order 3 if the LCM of the lengths of its disjoint cycles is 3. The possible cycle structures using 6 elements are:
- A single 3-cycle and three 1-cycles (fixed points). Cycle structure: (3, 1, 1, 1). The LCM(3, 1, 1, 1) = 3.
- Two disjoint 3-cycles. Cycle structure: (3, 3). The LCM(3, 3) = 3.
These are the only possible structures because the sum of cycle lengths must equal 6.
Calculating Elements for Each Structure
Structure 1: (3, 1, 1, 1)
To find the number of elements with this structure, we need to:
- Choose 3 elements out of 6 to form the 3-cycle. This can be done in $\binom{6}{3}$ ways.
- Arrange these 3 elements into a cycle. The number of distinct 3-cycles is $(3-1)!$.
Calculation:
- Number of ways to choose 3 elements: $\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
- Number of ways to form a 3-cycle from these 3 elements: $(3-1)! = 2! = 2$.
- Total elements for this structure: $\binom{6}{3} \times (3-1)! = 20 \times 2 = 40$.
Structure 2: (3, 3)
To find the number of elements with this structure, we need to:
- Partition the 6 elements into two groups of 3.
- Form a 3-cycle within each group.
Calculation:
- Number of ways to partition 6 elements into two sets of 3: We choose 3 elements for the first cycle ($\binom{6}{3}$) and the remaining 3 form the second cycle ($\binom{3}{3}$). Since the two cycles have the same length, the order of the sets doesn't matter, so we divide by $2!$. Number of partitions = $\frac{\binom{6}{3} \binom{3}{3}}{2!} = \frac{20 \times 1}{2} = 10$.
- Number of ways to form a 3-cycle from the first group of 3 elements: $(3-1)! = 2$.
- Number of ways to form a 3-cycle from the second group of 3 elements: $(3-1)! = 2$.
- Total elements for this structure: (Number of partitions) $\times (3-1)! \times (3-1)! = 10 \times 2 \times 2 = 40$.
Total Number of Elements of Order 3
Sum the counts from both possible structures:
Total = (Elements of type (3, 1, 1, 1)) + (Elements of type (3, 3))
Total = $40 + 40 = 80$.
Therefore, there are 80 elements of order 3 in the symmetric group $S_6$.
Important Questions from Permutations and Combinations
Two wizards try to create a spell using all the four elements, water, air, fire, and earth. For this, they decide to mix all these elements in all possible orders. They also decide to work independently. After trying all possible combination of elements, they conclude that the spell does not work.
How many attempts does each wizard make before coming to this conclusion, independently?A police station has six personnel comprising an inspector (I) and five constables ($C_1, C_2, C_3, C_4, C_5$). Normally, a 'raid team' of two, three, or four members is formed depending upon the nature of the raid. As per rule, it is mandatory that the inspector is part of raid team.
Number of distinct raid teams that can be formed is __________________.
(Answer in integer)
Five teams have to compete in a league, with every team playing every other team exactly once, before going to the next round. How many matches will have to be held to complete the league round of matches?
- How many 4-digit positive integers divisible by 3 can be formed using only the digits {1, 3, 4, 6, 7}, such that no digit appears more than once in a number?
- How many five-digit numbers can be formed using the integers 3, 4, 5 and 6 with exactly one digit appearing twice?
