A) \[\frac{6!8!10!}{4!6!}\]
B) \[\frac{8!8!10!}{4!6!}\]
C) \[\frac{8!8!6!}{6!4!}\]
D) None of these
Correct Answer: B
Solution :
[b] There are 8 chairs on each side of the table let the side be represented by A and B, Let four persons sit on side A, then number of ways of arranging 4 persons on 8 chairs on side \[A{{=}^{8}}{{P}_{4}}\] and then two persons sit on side B. The number of ways of arranging 2 persons on 8 chairs o side \[B{{=}^{8}}{{P}_{2}}\] and the remaining 10 persons can be arranged in remaining 10 chairs in 10! ways. Hence, the total number of ways in which the persons can be arranged \[{{=}^{8}}{{P}_{4}}{{\times }^{8}}{{P}_{2}}\times 10!=\frac{8!8!10!}{4!6!}\]
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