No. of four-digit even numbers formed by digits 0 1 2 3 4 5 without repetition I started from the units.
There can only be $3$ possibilities $0,2,4$.
Then there are $5$ possibilities for the tenth place.
THen there are $4$ possibilities for the hundredth place.
Then only $3$ possibilities remain.
Thus the permutation is $3 \times 4 \times 5\times 3 = 180$.
Is it correct?
Need guidance.
4 digit even numbers(including $0$ at the beginning) = $3* 5* 4* 3 = 180$   

for choosing the last digit we have 3 ways {0,2,4}. Now we are left with 5 numbers, for rest 3 places  $5*4*3$ ways

Now we will subtract those 4 digit even numbers with $0$ as the first digit = $2 * 4 * 3= 24$.
First digit is fixed = 0, 1 way. Now we are left with 2 choices for last digit (2,4). For choosing last digit 2 ways.
Now we have 4 digits and 2 places, $4*3$ ways.

Req.number of numbers = $180 - 24 = 156$
case(1): If repetition is not allowed,
1’s place can be filled by the nos (0,2 or 4) in 3P1 ways = 3 ways
10’s, 100’s and 1000’s places can be filled by 5 x 4 x 3 = 60 ways
Therefore, total number of arrangements = 60 x 3 = 180.
If ‘0’ comes in the 1000’s place, the number becomes 3 digits. Such 3 digits numbers =1 x 3 x 4 x 2 = 24.
Total number of 4 digit even numbers = 180  24 =156.