1. We need the number of four-digit numbers greater than $4321$ formed from digits $\{0,1,2,3,4,5\}$, with repetition allowed.

2. Total number of four-digit numbers possible:

   • First digit can be $1,2,3,4,5$ (cannot be $0$), so $5$ choices.
   • Each of the remaining three digits can be any of $0,1,2,3,4,5$, so $6^3$ choices.

   Hence, $5\cdot 6^3 = 5\cdot 216 = 1080.$

3. Instead of counting numbers greater than $4321$ directly, count numbers less than or equal to $4321$, then subtract from $1080$.

Count numbers $\le 4321$

We count by first digit.

Case 1: First digit is $1,2,$ or $3$

Then the number is automatically less than $4321$.

• Choices for first digit: $3$
• Remaining three digits: $6^3 = 216$

So count is $3\cdot 216 = 648.$

Case 2: First digit is $4$

Now we must ensure the number is $\le 4321$.

So consider numbers of form $4abc$.

Subcase 2.1: Second digit $<3$

Second digit can be $0,1,2$ : $3$ choices. Then last two digits are arbitrary: $6^2=36$.

Count: $3\cdot 36 = 108.$

Subcase 2.2: Second digit $=3$

Now number is $43bc$. We need $43bc \le 4321$.

Subcase 2.2.1: Third digit $<2$

Third digit can be $0,1$ : $2$ choices. Last digit arbitrary: $6$ choices.

Count: $2\cdot 6 = 12.$

Subcase 2.2.2: Third digit $=2$

Now number is $432d$ and we need $432d \le 4321$. So $d \le 1$, i.e. $d=0,1$ : $2$ choices.

Count: $2.$

Thus, with first digit $4$: $108 + 12 + 2 = 122.$

4. Therefore total numbers $\le 4321$ are $648 + 122 = 770.$

5. Hence numbers strictly greater than $4321$ are $1080 - 770 = 310.$

6. Therefore the correct option is $\boxed{310}.$ So, Option C is correct.