<s>[INST] 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.
 [/INST] You can approach this thinking that you have two different subsets: the even numbers $\{0,2,4\}$ and the odd numbers $\{1,3,5\}$. In order to avoid adjacent even numbers you have the two following cases:
Alternating even and odd numbers, starting with an odd number:
$3 \cdot 3 \cdot 2 \cdot 2 = 36$ possible combinations
First you can place any of the digits from $\{1,3,5\}$, then any from $\{0,2,4\}$, in the third place any of the remaining digits of the odd numbers set (2 in total), and same case for the even numbers in the last digit.
Alternating even and odd numbers, starting with an even number:
$2 \cdot 3 \cdot 2 \cdot 2 = 24$ possible combinations
The trick here is that starting with $0$ doesn't lead to a 4 digit number (you can add the zeros you want to the left  of a number and it remains the same number).
For even numbers at both extremes:
$2 \cdot 3 \cdot 2 \cdot 2 = 24$
For only one even number:
$2 \cdot 3 \cdot 2 \cdot 1 = 12$
$3 \cdot 3 \cdot 2 \cdot 1 = 18$
$3 \cdot 2 \cdot 3 \cdot 1 = 18$
$3 \cdot 2 \cdot 1 \cdot 3 = 18$
TOTAL: 
Therefore, in total, there are $(24+36+18\cdot 3+12+24=150)$ possibilities 
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