since the numbers need to be bigger then 500 the first digits can be 5 or 6 and the number can have 3, 4 or 5 digits (is it has only 2 digits will be smaller then 500)

- the first digit is 5 the rests of the digits are elements of {2, 3, 4, 6}

the total number of such numbers will be

4P2 + 4P3 + 4P4 = 4!((4-2)! + 4!/(4-3)! + 4!/(4-4)! = 2*3*4/2 + 2*3*4/1 + 2*3*4/1 = 12 + 24 + 24 = 60

where nPk = n!/(n-k)! where n! = 1*2*3*...*n, 0! = 1

- the first digit is 6 the rests of the digits are elements of {2, 3, 4, 5}

the total number of such numbers will be

4P2 + 4P3 + 4P4 = 4!((4-2)! + 4!/(4-3)! + 4!/(4-4)! = 2*3*4/2 + 2*3*4/1 + 2*3*4/1 = 12 + 24 + 24 = 60

thus the total number of such numbers will be 60 + 60 = 120.

there are 120 numbers greater than 500 and formed by choosing from the digits 2,3,4,5 and 6 without repeating of digits.