How many unique $3$-letter sequences with no spaces can be made using the letters in "AUGUSTIN LOUIS CAUCHY", which contains $19$ letters? For example, "GAA" is one acceptable sequence, but "GGA" is not an acceptable sequence because there is only one G available. The original ordering of the letters does not have to be preserved.

$\text{(A) }276\qquad\text{(B) }295\qquad\text{(C) }1486\qquad\text{(D) }1651\qquad\text{(E) }8086$
To solve this problem, we need to count the number of unique 3-letter sequences that can be formed using the letters in "AUGUSTIN LOUIS CAUCHY" without repeating any letter more than it appears in the original set. The original set contains 19 letters, but we need to consider the frequency of each letter.

First, let's list the letters and their frequencies:
- A: 2
- U: 3
- G: 1
- S: 2
- T: 1
- I: 2
- N: 1
- L: 1
- O: 1
- C: 1
- H: 1
- Y: 1

We will consider three cases: all letters are the same, two different letters, and three different letters.

1. **Case 1: All letters are the same**
   - We need to choose a letter that appears at least 3 times. The only letter that appears at least 3 times is 'U'.
   - Therefore, there is only 1 way to form such a sequence: "UUU".

2. **Case 2: Two different letters are used**
   - We need to choose 2 different letters where at least one of them appears twice.
   - We can choose the letter that appears twice or more in 3 ways: 'A', 'S', 'I'.
   - For each of these letters, we can pair it with any of the remaining 11 letters (since we are forming a 3-letter sequence, the order matters).
   - For each pair, we can arrange the letters in $\binom{3}{1} = 3$ ways (choosing 1 position for the letter that appears twice).
   - Therefore, the number of ways is:
     \[
     3 \times 11 \times 3 = 99
     \]

3. **Case 3: Three different letters are used**
   - We need to choose 3 different letters from the 12 unique letters.
   - The number of ways to choose 3 different letters from 12 is given by the combination:
     \[
     \binom{12}{3} = \frac{12!}{3!(12-3)!} = 220
     \]
   - For each combination of 3 letters, we can arrange them in $3! = 6$ ways.
   - Therefore, the number of ways is:
     \[
     220 \times 6 = 1320
     \]

Adding up all the cases, we get:
\[
1 + 99 + 1320 = 1486
\]

The final answer is $\boxed{1486}$