what is the number of elements of order 5 in the permutaion group S7?? so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7 to themselves....so we end up getting 2 cycles of length 1 each. the remaining five elements can be arranged in 4! ways... so, the answer is 7C2 * 4! = 21*24 = 504. but unfortunately, this answer is WRONG!! CAN ANYBODY TELL ME WHY?! PLEEEEASE HELP!