Question

The number of permutation of the letters of the word $HINDUSTAN$ such that neither the pattern ${}^{\prime }HI{N}^{\prime }$ nor ${}^{\prime }DU{S}^{\prime }$ nor ${}^{\prime }TA{N}^{\prime }$ appears, are :

• A. $166674$
• B. $169194$
• C. $166680$
• D. $181434$

Answer 1

Answer: (B)

$169194$

Total number of letters $=9$ in which N is repeated twice

Therefore total number of permutation $=\frac{9!}{2}$

The number of permutation in which HIN comes as block $=9-3+1=7!$

The number of permutation in which DUS comes as block $=7!$

The number of permutation in which TAN comes as block $=\frac{7!}{2}$

The number of permutation in which HIN and DUS comes as block $=5!$

The number of permutation in which DUS and TAN comes as block $=5!$

The number of permutation in which HIN and TAN comes as block $=5!$

Therefore the required number of permutations $=\frac{9!}{2}-(7!+7!+\frac{7!}{2}-3\times 5!+3!)$

$=\frac{362880}{2}-(5040+5040+\frac{5040}{2}-360+6)$

$=181440-(10080+2520-360+6)$

$=181440-12246$

$=169194$