To solve the problem of finding the number of permutations of the letters in the word "HINDUSTAN" such that none of the patterns 'HIN', 'DUS', or 'TAN' appear, we can follow these steps:
### Step 1: Count the total number of letters
The word "HINDUSTAN" consists of 9 letters. The letters are: H, I, N, D, U, S, T, A, N. Here, the letter 'N' is repeated twice.
### Step 2: Calculate the total permutations
The total number of permutations of the letters in the word can be calculated using the formula for permutations of multiset:
\[
\text{Total permutations} = \frac{n!}{p_1! \cdot p_2! \cdots}
\]
where \(n\) is the total number of letters and \(p_i\) are the counts of each repeated letter.
For "HINDUSTAN":
\[
\text{Total permutations} = \frac{9!}{2!} = \frac{362880}{2} = 181440
\]
**Hint:** Remember to account for repeated letters when calculating permutations.
### Step 3: Count permutations including the patterns
Next, we need to count the permutations where the patterns 'HIN', 'DUS', or 'TAN' appear.
#### Case 1: 'HIN' as a block
If we treat 'HIN' as a single block, we have the blocks: HIN, D, U, S, T, A, N. This gives us 7 blocks in total (1 block of 'HIN' and 6 individual letters).
\[
\text{Permutations} = \frac{7!}{2!} = \frac{5040}{2} = 2520
\]
#### Case 2: 'DUS' as a block
Similarly, treating 'DUS' as a single block gives us the blocks: H, I, N, DUS, T, A, N. Again, we have 7 blocks.
\[
\text{Permutations} = \frac{7!}{2!} = 2520
\]
#### Case 3: 'TAN' as a block
Treating 'TAN' as a block gives us the blocks: H, I, N, D, U, S, TAN. Again, we have 7 blocks.
\[
\text{Permutations} = \frac{7!}{2!} = 2520
\]
**Hint:** When treating a group of letters as a block, reduce the total count of letters accordingly.
### Step 4: Count permutations including two patterns
Now, we need to consider cases where two patterns appear together.
#### Case 4: 'HIN' and 'DUS' as blocks
We treat 'HIN' and 'DUS' as blocks, giving us: HIN, DUS, T, A, N (5 blocks).
\[
\text{Permutations} = \frac{5!}{2!} = \frac{120}{2} = 60
\]
#### Case 5: 'HIN' and 'TAN' as blocks
We treat 'HIN' and 'TAN' as blocks, giving us: HIN, D, U, S, TAN (5 blocks).
\[
\text{Permutations} = \frac{5!}{2!} = 60
\]
#### Case 6: 'DUS' and 'TAN' as blocks
We treat 'DUS' and 'TAN' as blocks, giving us: H, I, N, DUS, TAN (5 blocks).
\[
\text{Permutations} = \frac{5!}{2!} = 60
\]
**Hint:** When two patterns are treated as blocks, reduce the total count of letters accordingly.
### Step 5: Count permutations including all three patterns
Finally, if 'HIN', 'DUS', and 'TAN' are all treated as blocks, we have: HIN, DUS, TAN (3 blocks).
\[
\text{Permutations} = 3! = 6
\]
### Step 6: Apply the Inclusion-Exclusion Principle
Now we apply the Inclusion-Exclusion Principle to find the total number of permutations that contain at least one of the patterns:
\[
\text{Total with patterns} = (2520 + 2520 + 2520) - (60 + 60 + 60) + 6
\]
\[
= 7560 - 180 + 6 = 7386
\]
### Step 7: Calculate the required permutations
Finally, we subtract the number of permutations that include at least one of the patterns from the total permutations:
\[
\text{Required permutations} = 181440 - 7386 = 174054
\]
### Final Answer
The number of permutations of the letters of the word "HINDUSTAN" such that neither the pattern 'HIN', 'DUS', nor 'TAN' appears is **174054**.
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