A) 166674
B) 169194
C) 166680
D) 181434
Correct Answer: B
Solution :
[b] Total number of permutations \[=\frac{9!}{2!}\] |
Number of those containing \['HIN'\] \[=7!\] |
Number of those containing ?DUS? \[=\frac{7!}{2!}\] |
Number of those containing ?TAN? \[=7!\] |
Number of those containing ?HIN? and ?DUS? \[=5!\] |
Number of those containing ?HIN? and ?TAN?\[=5!\] |
Number of those containing ?TAN? and ?DUS? \[=5!\] |
Number of those containing ?HIN? and ?DUS? and ?TAN?=3! |
Required number |
\[=\frac{9!}{2!}-\left( 7!+7!+\frac{7!}{2} \right)+3\times 5!-3!=169194.\] |
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