A) \[9.7!\]
B) \[8!\]
C) \[5!.4!\]
D) \[8!.4!\]
Correct Answer: A
Solution :
Number of digits are\[9\] Select \[2\] places for the digit \[1\] and \[2\] in \[^{9}{{C}_{2}}\] ways from the remaining \[7\] places select any two places for \[3\] and \[4\] in \[^{7}{{C}_{2}}\] ways and from the remaining \[5\] places select any two for \[5\] and \[6\] in \[^{5}{{C}_{2}}\] ways Now, the remaining 3 digits can be filled in \[3!\] ways \[\therefore \]Total ways\[{{=}^{9}}{{C}_{2}}{{\cdot }^{7}}{{C}_{2}}{{\cdot }^{5}}{{C}_{2}}\cdot 3!\] \[=\frac{9!}{2!.7!}\cdot \frac{7!}{2!.5!}\cdot \frac{5!}{2!3!}=9.7!\]You need to login to perform this action.
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