Question

The number of permutations of the letters of the word HINDUSTAN such that neither the pattern ‘HIN’ nor ‘DUS’ nor ‘TAN’ appears, are:

Answer 1

Hint: We are given a word HINDUSTAN. Firstly, find the total number of permutations of the letters of the word HINDUSTAN.
Since we need to find the number of permutations of the letters of the word HINDUSTAN such that neither the pattern ‘HIN’ nor ‘DUS’ nor ‘TAN’ appears. So, find the number of permutations of words containing HIN, DUS, and TAN respectively. Then find the number of permutations of words containing HIN & DUS, DUS & TAN, and HIN &TAN respectively. Also, find the number of permutations containing HIN, DUS, and TAN altogether. Add all the permutations of words containing HIN, DUS, and TAN respectively, and permutations containing HIN, DUS, and TAN altogether and subtract permutations of words containing HIN & DUS, DUS & TAN, and HIN &TAN respectively. Then subtract the answer from the total number of permutations of the letters of the word HINDUSTAN. You get the answer required.

Complete step-by-step answer:
As we are given the word HINDUSTAN. In the word HINDUSTAN, the total number of letters = 9 out of which N repeats twice.
So, the total ways of permutations of the letters of the word HINDUSTAN are:
$\begin{align}
  & =\dfrac{9!}{2!} \\
 & =181440 \\
\end{align}$
Now, find the number of permutations of words containing HIN, where HIN is considered as one letter altogether and the remaining 6 places are to be filled with other 6 letters.
So, we have 7 places and 7 letters. Therefore, the number of permutations of words containing HIN together is:
$\begin{align}
  & =7! \\
 & =5040 \\
\end{align}$
Similarly, the permutations of TAN would be:
$\begin{align}
  & =7! \\
 & =5040 \\
\end{align}$
Now, for the permutations of DUS, we have 7 letters but N is being repeated twice. So, the permutation of words containing DUS would be:
$\begin{align}
  & =\dfrac{7!}{2!} \\
 & =2520 \\
\end{align}$
Now, find the permutations of words containing HIN and DUS together, where HIN and DUS are considered as two letters and the remaining 3 places are to be filled with other 3 letters.
So, we have 5 places and 5 letters. Therefore, the number of permutations of words containing HIN together is:
$\begin{align}
  & =5! \\
 & =120 \\
\end{align}$
Similarly, permutations for words containing TAN and DUS would be: 120
And permutations for words containing HIN and TAN would be: 120
Now, the total number of permutations for words containing HIN, DUS, and TAN where HIN, DUS, and TAN are considered as 3 different letters.
So, we have 3 places and 3 letters. Therefore, the number of permutations of words containing HIN, DUS and TAN together is:
$\begin{align}
  & =3! \\
 & =6 \\
\end{align}$

Now, the number of permutations of the letters of the word HINDUSTAN such that neither the pattern ‘HIN’ nor ‘DUS’ nor ‘TAN’ appears is:
$\begin{align}
  & =181440-\left( 5040+5040+2520-120-120-120+6 \right) \\
 & =169194 \\
\end{align}$

Note: Since the letter N is repeating twice. So, do not forget to divide by $2!$, wherever required. Otherwise, you will end up with the wrong number of permutations of the words formed.