A quarry wants to sell a large pile of gravel. At full price, the gravel would sell for $3200$ dollars. But during the first week the quarry only sells $60\%$ of the gravel at full price. The following week the quarry drops the price by $10\%$, and, again, it sells $60\%$ of the remaining gravel. Each week, thereafter, the quarry reduces the price by another $10\%$ and sells $60\%$ of the remaining gravel. This continues until there is only a handful of gravel left. How many dollars does the quarry collect for the sale of all its gravel?
1. **First Week:**
   - The quarry sells $60\%$ of the gravel at full price.
   - Full price of the gravel is $3200$ dollars.
   - Amount sold in the first week: $0.6 \times 3200 = 1920$ dollars.

2. **Second Week:**
   - The remaining gravel is $40\%$ of the original amount.
   - The price is reduced by $10\%$, so the new price is $90\%$ of the original price.
   - New price: $0.9 \times 3200 = 2880$ dollars.
   - Amount sold in the second week: $0.6 \times 0.4 \times 2880 = 0.6 \times 0.4 \times 3200 \times 0.9 = 691.2$ dollars.

3. **Third Week:**
   - The remaining gravel is $40\%$ of the previous remaining amount, which is $40\%$ of $40\%$ of the original amount.
   - The price is reduced by another $10\%$, so the new price is $90\%$ of the previous price.
   - New price: $0.9 \times 0.9 \times 3200 = 0.81 \times 3200 = 2592$ dollars.
   - Amount sold in the third week: $0.6 \times 0.4^2 \times 2592 = 0.6 \times 0.4^2 \times 3200 \times 0.9^2 = 248.832$ dollars.

4. **General Term:**
   - For the $n$-th week, the remaining gravel is $0.4^{n-1}$ of the original amount.
   - The price is reduced by $10\%$ each week, so the price is $0.9^{n-1}$ of the original price.
   - Amount sold in the $n$-th week: $0.6 \times 0.4^{n-1} \times 0.9^{n-1} \times 3200$ dollars.

5. **Sum of the Infinite Series:**
   - The series is geometric with the first term $a = 1920$ and common ratio $r = 0.36$.
   - The sum of an infinite geometric series is given by $\frac{a}{1 - r}$.
   - Here, $a = 1920$ and $r = 0.36$.
   - Sum: $\frac{1920}{1 - 0.36} = \frac{1920}{0.64} = 3000$ dollars.

\[
\sum_{n=1}^{\infty} 0.6 \times 0.4^{n-1} \times 0.9^{n-1} \times 3200 = \frac{1920}{1 - 0.36} = 3000
\]

The final answer is $\boxed{3000}$