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.
To solve the problem of finding 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, we can use a combinatorial approach. Let's denote the places as $A, B, C, D$.

1. **Total number of 4-digit numbers:**
   - The first digit $A$ can be any of $\{1,2,3,4,5\}$ (5 choices, excluding 0).
   - The second digit $B$ can be any of the remaining 5 digits (including 0).
   - The third digit $C$ can be any of the remaining 4 digits.
   - The fourth digit $D$ can be any of the remaining 3 digits.
   
   Therefore, the total number of 4-digit numbers is:
   \[
   5 \times 5 \times 4 \times 3 = 300
   \]

2. **Counting the complement (numbers with two adjacent even digits):**
   - Even digits: $\{0, 2, 4\}$
   - Odd digits: $\{1, 3, 5\}$

   We will consider different cases where two adjacent digits are even.

   **Case I:** $A, B$ are even, $C, D$ are odd.
   - $A$ can be $\{2, 4\}$ (2 choices).
   - $B$ can be $\{0, 2, 4\} - \{A\}$ (2 choices).
   - $C, D$ can be any two numbers from $\{1, 3, 5\}$ (3 choices for $C$, 2 choices for $D$).
   
   Total possibilities: 
   \[
   2 \times 2 \times 3 \times 2 = 24
   \]

   **Case II:** $B, C$ are even, $A, D$ are odd.
   - $A$ can be any of $\{1, 3, 5\}$ (3 choices).
   - $B$ can be $\{0, 2, 4\}$ (3 choices).
   - $C$ can be $\{0, 2, 4\} - \{B\}$ (2 choices).
   - $D$ can be any of the remaining odd digits (2 choices).
   
   Total possibilities:
   \[
   3 \times 3 \times 2 \times 2 = 36
   \]

   **Case III:** $C, D$ are even, $A, B$ are odd.
   - This case is similar to Case II.
   
   Total possibilities:
   \[
   3 \times 3 \times 2 \times 2 = 36
   \]

   **Case IV:** $A, B, C$ are even, $D$ is odd.
   - $A$ can be $\{2, 4\}$ (2 choices).
   - $B$ can be $\{0, 2, 4\} - \{A\}$ (2 choices).
   - $C$ can be $\{0, 2, 4\} - \{A, B\}$ (1 choice).
   - $D$ can be any of the odd digits (3 choices).
   
   Total possibilities:
   \[
   2 \times 2 \times 1 \times 3 = 12
   \]

   **Case V:** $B, C, D$ are even, $A$ is odd.
   - $A$ can be any of $\{1, 3, 5\}$ (3 choices).
   - $B$ can be $\{0, 2, 4\}$ (3 choices).
   - $C$ can be $\{0, 2, 4\} - \{B\}$ (2 choices).
   - $D$ can be $\{0, 2, 4\} - \{B, C\}$ (1 choice).
   
   Total possibilities:
   \[
   3 \times 3 \times 2 \times 1 = 18
   \]

   **Case VI:** $A, C, D$ are even, $B$ is odd.
   - $A$ can be $\{2, 4\}$ (2 choices).
   - $B$ can be any of $\{1, 3, 5\}$ (3 choices).
   - $C$ can be $\{0, 2, 4\} - \{A\}$ (2 choices).
   - $D$ can be $\{0, 2, 4\} - \{A, C\}$ (1 choice).
   
   Total possibilities:
   \[
   2 \times 3 \times 2 \times 1 = 12
   \]

   **Case VII:** $A, B, D$ are even, $C$ is odd.
   - This case is similar to Case VI.
   
   Total possibilities:
   \[
   2 \times 3 \times 2 \times 1 = 12
   \]

   Summing up all the cases:
   \[
   24 + 36 + 36 + 12 + 18 + 12 + 12 = 150
   \]

3. **Finding the desired count:**
   - Total 4-digit numbers: 300
   - Numbers with two adjacent even digits: 150
   
   Therefore, the number of 4-digit numbers with no two adjacent even digits is:
   \[
   300 - 150 = 150
   \]

The final answer is $\boxed{150}$.