How many $4$-digit even numbers can be formed from a certain set? Let's imagine a set with $6$ numbers, of which $3$ are even (including $0$) and $3$ are odd. How many $4$-digit even numbers can be formed? Digits cannot be repeated.
Zero is present: $5 \cdot 4 \cdot 3 \cdot 1$
Zero is not present: $4 \cdot 3 \cdot 2 \cdot 2$
Summing both outputs gives $108$ cases, but it is not correct.

We have $4$ positions, $p_{1},p_{2},p_{3},p_{4}$. 


*

*We can put $0$ in $p_{4}$. In this case, we have $P(5,3)$ for remaining digits.

*We can put $0$ in $p_{2}$ or $p_{3}$. In this case, $2$ even digits remains and we must put one of them in $p_{4}$.Thus, $C(2,1) \times C(2,1) \times P(4,2)$.

*We can have a number without $0$. In this case, $C(2,1)\times P(4,3)$. 


Therefore, the result is $$P(5,3) + C(2,1) \times C(2,1) \times P(4,2) + C(2,1)\times P(4,3)$$
