Find the number of numbers formed by 0, 2, 4, 6, 8 that are > 4000 without repetition

Given digits are 0,2, 4, 6,8 The number of numbers greater than 4000:-
Case (i)
All the 5 digited numbers are greater than 4000 In 5 digited numbers zero cannot be filled in the first place we can fill the first place with the remaining 4 digits in 4 ways and the remaining 4 places can be filled with the remaining 4 digits in 4! ways

$\therefore$ The no. of 5 digited numbers = 4×4! = 96

Case (ii)
In the 4 digit numbers, the number starting with 4 or 6 or 8 are greater than 4000. It can be done in 3 ways and the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ ways

$\therefore$The number of 4 digited number greater than
4000 $=$ 3 $\times$⁴P₃ =72
$\therefore$ Req. no. of arrangements =96+72 = 168 ways