求从单词 ENDANGERED 中，在下列情况下可以组成 4 字母密码字的方案数：

1. Identify the set of distinct letters available in the word

The word ENDANGERED contains the letters E, N, D, A, G, R.
There are 6 distinct letters to choose from.

2. Calculate the number of 4-letter arrangements with no repeated letters

Since order matters for code-words and there are no repeated letters, the number of ways is:
$^6P_4 = 6 \times 5 \times 4 \times 3 = 360$

1. Determine the number of choices for the remaining letter.

The word ENDANGERED consists of the letters E(3), N(2), D(2), A(1), G(1), R(1).
Since the code-word must contain exactly three "E"s, we need to choose 1 more letter from the $5$ remaining distinct letters {N, D, A, G, R}.
Number of choices for the final letter $= \binom{5}{1} = 5$.

2. Determine the number of valid arrangements and calculate the total.

For any chosen set of $4$ letters (e.g., E, E, E, N), the number of distinct arrangements with $3$ identical items is $\frac{4!}{3!} = 4$.
Total number of code-words $= 5 \times 4 = 20$.

1. Identify the possible cases for having at least one repeated letter

The multiset of letters from ENDANGERED is {3E, 2N, 2D, 1A, 1G, 1R}.
To find the number of ways with at least one repeated letter, we must split the problem into mutually exclusive cases based on the type of repetition:

1. Exactly one pair of identical letters (e.g., AABC)
2. Two pairs of identical letters (e.g., AABB)
3. Three identical letters (e.g., AAAB)

2. Calculate the number of ways to form a word with exactly one pair of identical letters

Case 1: Exactly one pair of identical letters
Choose $1$ letter to be the pair from {E, N, D}: $3$ choices.
Choose $2$ distinct letters from the remaining $5$ letter types: $^{5}C_2 = 10$ choices.
Number of permutations for each selection: $\frac{4!}{2!} = 12$.
Number of ways = $3 \times 10 \times 12 = 360$.

3. Calculate the number of ways to form a word with two pairs of identical letters

Case 2: Two pairs of identical letters
Choose $2$ letters to be the pairs from {E, N, D}: $^3C_2 = 3$ choices.
Number of permutations for each selection: $\frac{4!}{2!2!} = 6$.
Number of ways = $3 \times 6 = 18$.

4. Incorporate the case of three identical letters and sum all valid cases

Case 3: Three identical letters
From part (ii), the number of words with three 'E's and $1$ distinct letter is $20$ ways.
(Working: $1 \text{ (choose E)} \times ^{5}C_1 \text{ (choose other letter)} \times \frac{4!}{3!} = 20$)

By the addition principle, the total number of ways with at least one repeated letter is:
$\text{Total} = 360 + 18 + 20 = 398$.