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# True-or-False:Eighteen guests are to be seated,half on each side of a long table.Four particular guests desire to sit on one particular side and three others on other side of the table.The number of ways in which the seating arrangements can be made is $\large\frac{11!}{5!6!}$$(9!)(9!) \begin{array}{1 1}(A)\;\text{True}\\(B)\;\text{False}\end{array} Can you answer this question? ## 1 Answer 0 votes We can select 5 of them in 11C_5 ways to sit along with 4. Rest will automatically go to the other side .This can be done only way Now 9 on either side can be arranged in 9! ways So the total number of arrangements are =11C_5\times 9!\times 9! \Rightarrow \large\frac{11!}{5!6!}$$\times 9!\times 9!$
Hence the given statement is true.
edited May 16, 2014