If letters of the word “THING” are arranged in all possible manner and words thus formed are written in dictionary order. If K is the number of words lying between “NIGHT” and “THING” (both exclusive) in that dictionary, then:

Options

A. Number of zeros at the end of K! is 4
B. Number of divisors of K is 6
C. Number of integral coordinates (both abacissa and ordinate integer) lying strictly inside triangle formed by y=0, x=0, x+y=K is 171
D. K does not divide number of words in that dictionary

Answer

To solve the problem, we need to find the number of words that lie between "NIGHT" and "THING" when all permutations of the letters in "THING" are arranged in dictionary order.

### Step 1: List the letters in alphabetical order
The letters of the word "THING" are: T, H, I, N, G.
Arranging these letters in alphabetical order gives us:
- G, H, I, N, T

### Step 2: Calculate the total permutations
The total number of permutations of the letters in "THING" is given by the formula for permutations of n distinct objects, which is \( n! \). Here, \( n = 5 \) (since there are 5 letters).

\[ 5! = 120 \]

### Step 3: Count the words before "NIGHT"
To find the number of words that come before "NIGHT", we will consider the permutations starting with each letter that comes before 'N' in the alphabetical order.

1. **Starting with G**: The remaining letters are H, I, N, T. The number of permutations is:
\[ 4! = 24 \]

2. **Starting with H**: The remaining letters are G, I, N, T. The number of permutations is:
\[ 4! = 24 \]

3. **Starting with I**: The remaining letters are G, H, N, T. The number of permutations is:
\[ 4! = 24 \]

4. **Starting with N**: Now we need to consider the permutations starting with 'N' and the next letter must be 'I' to get to "NIGHT".

- **N I**: The remaining letters are G, H, T. The number of permutations is:
\[ 3! = 6 \]

- **N H**: The remaining letters are G, I, T. The number of permutations is:
\[ 3! = 6 \]

- **N G**: The remaining letters are H, I, T. The number of permutations is:
\[ 3! = 6 \]

- **N I G**: The remaining letters are H, T. The number of permutations is:
\[ 2! = 2 \]

- **N I H**: The remaining letters are G, T. The number of permutations is:
\[ 2! = 2 \]

- **N I T**: The remaining letters are G, H. The number of permutations is:
\[ 2! = 2 \]

Now, we can sum all these permutations to find the total number of words before "NIGHT":
\[ 24 + 24 + 24 + 6 + 6 + 6 + 2 + 2 + 2 = 96 \]

### Step 4: Count the words before "THING"
Now we need to count the words before "THING".

1. **Starting with G**: The remaining letters are H, I, N, T. The number of permutations is:
\[ 4! = 24 \]

2. **Starting with H**: The remaining letters are G, I, N, T. The number of permutations is:
\[ 4! = 24 \]

3. **Starting with I**: The remaining letters are G, H, N, T. The number of permutations is:
\[ 4! = 24 \]

4. **Starting with N**: The remaining letters are G, H, I, T. The number of permutations is:
\[ 4! = 24 \]

5. **Starting with T**: We need to consider the permutations starting with 'T':
- **T G**: The remaining letters are H, I, N. The number of permutations is:
\[ 3! = 6 \]

- **T H**: The remaining letters are G, I, N. The number of permutations is:
\[ 3! = 6 \]

- **T I**: The remaining letters are G, H, N. The number of permutations is:
\[ 3! = 6 \]

- **T N**: The remaining letters are G, H, I. The number of permutations is:
\[ 3! = 6 \]

Now, we can sum all these permutations to find the total number of words before "THING":
\[ 24 + 24 + 24 + 24 + 6 + 6 + 6 + 6 = 120 \]

### Step 5: Calculate K
Now we can find K, which is the number of words lying between "NIGHT" and "THING" (both exclusive):
\[ K = \text{(Number of words before "THING")} - \text{(Number of words before "NIGHT")} - 1 \]
\[ K = 120 - 96 - 1 = 23 \]

### Final Answer
Thus, the value of K is 23.