How many zeros are there in the last digits of the following number $P = 11\times12\times ...\times 88\times 89$ ?

(A): $16$, (B): $17$, (C): $18$, (D): $19$, (E) None of the above.
To determine the number of zeros in the last digits of the product \( P = 11 \times 12 \times \cdots \times 88 \times 89 \), we need to count the number of factors of 10 in \( P \). A factor of 10 is composed of a factor of 2 and a factor of 5. Since there are generally more factors of 2 than factors of 5 in a factorial, we only need to count the number of factors of 5.

1. **Using Legendre's Formula**:
   Legendre's formula for the exponent of a prime \( p \) in \( n! \) is given by:
   \[
   e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
   \]
   We need to apply this formula to both \( 89! \) and \( 10! \) to find the number of factors of 5 in each.

2. **Counting factors of 5 in \( 89! \)**:
   \[
   e_5(89!) = \left\lfloor \frac{89}{5} \right\rfloor + \left\lfloor \frac{89}{25} \right\rfloor + \left\lfloor \frac{89}{125} \right\rfloor + \cdots
   \]
   Calculating each term:
   \[
   \left\lfloor \frac{89}{5} \right\rfloor = 17, \quad \left\lfloor \frac{89}{25} \right\rfloor = 3, \quad \left\lfloor \frac{89}{125} \right\rfloor = 0
   \]
   Therefore:
   \[
   e_5(89!) = 17 + 3 = 20
   \]

3. **Counting factors of 5 in \( 10! \)**:
   \[
   e_5(10!) = \left\lfloor \frac{10}{5} \right\rfloor + \left\lfloor \frac{10}{25} \right\rfloor + \left\lfloor \frac{10}{125} \right\rfloor + \cdots
   \]
   Calculating each term:
   \[
   \left\lfloor \frac{10}{5} \right\rfloor = 2, \quad \left\lfloor \frac{10}{25} \right\rfloor = 0
   \]
   Therefore:
   \[
   e_5(10!) = 2
   \]

4. **Counting factors of 5 in \( P \)**:
   Since \( P = \frac{89!}{10!} \), the number of factors of 5 in \( P \) is:
   \[
   e_5(P) = e_5(89!) - e_5(10!) = 20 - 2 = 18
   \]

Conclusion:
The number of zeros in the last digits of \( P \) is \( \boxed{18} \).