A Newspaper Permuted

The possible permutations for HINDUSTAN is $\frac{9!}{2!} =181440$

Now find the possible permutations of consisting $'HIN'$ , $'DUS'$ , or $'TAN'$

1. The number of permutations of consisting $'HIN'$ is $7!$

2. The number of permutations of consisting $'DUS'$ is $7!/2!$ (since N are ordered twice)

3. The number of permutations of consisting $'TAN'$ is $7!$

Now there is a possibility that each case of $'HIN'$ , $'DUS'$ , or $'TAN'$ are arranged more than twice, such as $DUSHINNAT$ since this arrangement occur on the case either 1 or 2. Then, we make another case:

1. The number of permutations of consisting $'HIN'$ and $'DUS'$ is $5!$

2. The number of permutations of consisting $'DUS'$ and $'TAN'$ is $5!$

3. The number of permutations of consisting $'HIN'$ and $'TAN'$ is $5!$

Now we erase the doubles and we accidentally erase all the possibilities of consisting $'HIN'$ , $'DUS'$ and $'TAN'$ . The number of permutations of this case is $3!$

Finally the possible permutations of consisting $'HIN'$ , $'DUS'$ , or $'TAN'$ is $7!+\frac{7!}{2!} + 7! - 5!-5!-5!+3! = 12246$ possibilities.

Since we want to find the complement of the permutations, we have subtract the cases from the all possibilities. So we have $181440-12246=169194$ possibilities to make permutations with conditions neither consisting $'HIN'$ , $'DUS'$ nor $'TAN'$

For simply one, use inclusive and exclusive principle

Python 2.7:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

from itertools import permutations
def goal(test):
    return 'hin' not in test and\
       'dus' not in test and\
       'tan' not in test
found = set()
count = 0
for combo in permutations('hindustan'):
    test = ''.join(combo)
    if test not in found:
        found.add(test)
        if goal(test):
            count += 1
print "Answer:", count