Since all even digits need to be adjacent, we can first figure out how many ways we can arrange the even digits.
To do this, we start by fixing one digit in the front, and permute the next two. Then we bring the second to the front, fix, and permute the other two. This will get the following combinations:
(024), (042), (204), (240), (420), (402)
For each of these, we get 3 different positions to place them:
0 2 4 X Y Z X 0 2 4 Y Z X Y 0 2 4 Z X Y Z 0 2 4 Do the same this for the other 5 combinations.
**Note: exclude the numbers with led by 0. (0 2 4 X Y Z and 0 4 2 X Y Z)
For each combination, we have 4 different places to place them,
(number of combination) * (number of ways to place) => 6 * 4 = 24
24 - 2 = 22 to exclude the ones with leading 0
Now we need to think about how many different ways are there to pick X, Y, and Z
For each X, you have 3 possible digits,0, 1, or 5. After X is picked, you only have 2 digits left to pick for Y and then 1 digit left for Z. So for each X, Y, Z, we have 3! = 3 * 2 * 1 ways to pick the digits.
((Number of combination * number of ways to place the combinations) - the leading 0s) * ways to pick other digits, we get ((6 * 4) - 2) * 6 = 132