Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.

Since you have three odd and three even digits, you must use at least one even digit. But you cannot use three even digits as that would force two adjacent even digits.
a) One even digit: There are $3$ ways to choose the even digit, and $3\cdot 2\cdot 1$ ways to choose the odd digits from left to right, and $4$ ways to choose the position of the even one:

 $3\cdot 3\cdot 2\cdot 1\cdot 4=72$

b) Two even digits: There are $3\cdot 2$ ways to choose the even digits from left to right, and $3\cdot 2$ ways to choose the odd digits from left to right, and ${4\choose 2}-3$ ways to choose the even positions non-adjacent: 

$3\cdot 2\cdot 3\cdot 2\cdot(6-3)=108$

In total, we get

 $72+108=180$.

(The above assumes that leading $0$ is allowed.)
