How many positive integers less than $2010$ are there such that the sum of factorials of its digits is equal to itself?

$ \textbf{(A)}\ 5
\qquad\textbf{(B)}\ 4
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 2
\qquad\textbf{(E)}\ \text{None}
$
1. **Understanding the problem**: We need to find the number of positive integers less than 2010 such that the sum of the factorials of its digits equals the number itself.

2. **Constraints on digits**: Since $7! = 5040 > 2010$, no digit in the number can be greater than 6. Therefore, the digits of the number can only be 0, 1, 2, 3, 4, 5, or 6.

3. **Checking possible numbers**:
    - If a number has three digits, the maximum sum of factorials of its digits is $3 \cdot 6! = 3 \cdot 720 = 2160$, which is greater than 2010. Therefore, we need to check numbers with fewer digits.

4. **Analyzing numbers with different digits**:
    - **Three sixes**: $n > 3 \cdot 6! = 2160$ (impossible since $n < 2010$).
    - **Two sixes**: $n = 2 \cdot 6! + a! \leq 1440 + a! \leq 1680$. This implies $a$ must be 1, but no choice of $a$ produces a valid number.
    - **One six**: $720 < n < 720 + 2 \cdot 5! < 1000$. This implies $n$ would contain a digit greater than 6, which is not possible.

5. **Numbers without sixes**:
    - **Three fives**: $n = 3 \cdot 5! = 360$ (no valid number).
    - **Two fives**: $n = 2 \cdot 5! + a! \leq 240 + a! \leq 264$ (no valid number).
    - **One five**: $n = 5! + a! + b! = 120 + a! + b!$. Trying $a = 4$ and $b = 1$, we get $n = 120 + 24 + 1 = 145$, which is valid.

6. **Numbers without fives or sixes**:
    - **Three digits**: $n \leq 3 \cdot 4! = 72$.
    - **Two digits**: $n \leq 2 \cdot 4! = 48$. Checking numbers:
        - $1! = 1$
        - $2! = 2$
        - Other combinations (11, 12, 13, 14, 22, 23, 24, 33, 34, 44) do not work.

7. **Conclusion**: The valid numbers are 1, 2, and 145.

The final answer is $\boxed{3}$.