Show that the number of permutations of 3m for which all cyc

Show that the number of permutations of [3m] for which all cycles have lengths which are multiples of 3 is (3m - 1)(3m - 2)^2 (3m - 4)(3m - 5)^2 .... 2. 1^2.

Solution

Proof:- to prove this, let us recall some properties,

Let k greater than or equal to 2, the number of permutations all of whose cycle lengths are divisible by k is given by 1².2.3.......(k-1)(k+1)²(k+2)........(2k-1)(2k-1)²....(2k+2).......(n-1)

let k greater than or equal to 2 , the number of permutations , none of whose cycles of lengths is divisble by k is given by,

1.2.3........(k-1)²(k+1).....(2k-1)(2k-1)²(2k+1).......(n-1)² if k does not divides n.

1.2.3..........(k-1)²(k+1)........(2k-2)(2k-1)²(2k+1)......(n-2)(n-1)² if k divides n.

now, in the above problem, the cycle have lengths multiples of 3.

Now we have to find the number of permutations of [3m] , in which cycle of length, which are multiples of 3 is,(3m-1)(3m-2)²(3m-4)......2.1²

given that, the number of permutations [3m] which are multiples of 3, so leave the multiples of 3, that is leave the values of (3m-3)3m-6)..............3, we will get the required permutation permutation.

hence proved.