I have solved it as per my knowledge and understanding.
Since there are 4 elements so no. of permutations will be
$$4! = 24$$
$$(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (2,1,3,4), (2,1,4,3), (2,3,1,4), (2,3,4,1), (2,4,1,3), (2,4,3,1), (3,2,1,4), (3,2,4,1), (3,1,2,4), (3,1,4,2), (3,4,2,1), (3,4,1,2), (4,2,3,1), (4,2,1,3), (4,3,2,1), (4,3,1,2), (4,1,2,3), (4,1,3,2)$$
Is it correct?
• And for even and odd-
Even: product of even no of transpositions. e.g., $(1,2,3) = (1,2)(1,3)$ is even
Odd: product of odd no of transpositions. e.g., $(1,2,3,4) = (1,2)(1,3)(1,4)$ is odd.
• But if I apply this theory in this question then all permutations will be odd.
I am confused here, is my solution incorrect? May be there will be more permutations with less elements like $(1,2,3) , (1,2,4)$?
• Also there is a swapping logic! If the numbers are swapped odd times then it is odd and even otherwise
So in this case $(1,2,3,4)$ is even (no swaps)
$(3,2,1,4)$ = Swap 1<->3 is odd
Also $(4,2,1,3)$ = Swap 3<->4 then swap 1<->4 = even permutation
What will be the answer to my question?
Best Answer
Don't be put off! You have made a very simple mistake which might be easily rectified.
I haven't checked your list for completeness, but I suspect it does include all the permutations, you've just mixed up two different notations.
I assume for example that in the list $(4,2,3,1)$ is the permutation mapping $1\mapsto 4$, $2\mapsto 2$, $3\mapsto 3$, $4\mapsto 1$? The usual (or cycle) notation for this would be $(1,4)$, an odd permutation.
As you have written $(2,3,1,4)$, would usually be written $(1,2,3)=(1,2)(1,3)$ (or $(1,3)(1,2)$ depending on the convention you use), an even permutation.
See this wiki page for a brief description of the cycle notation.
As for how many are even and how many are odd, you must have proven that a cycle is either even or odd, not both. So if $x$ is an odd permutation, then it is the product of an odd number of transpositions, so $(1,2)x$ is a product of an even number of transpositions so is even. This map $x\mapsto (1,2)x$ is a bijection between the set of odd and the set of even permutations. What does this tell you about how much there must be of each?