Find the number of ways in which 4-letter code-words can be obtained from the word ENDANGERED if

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:
$\text{Total} = 360 + 18 + 20 = 398$.