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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} $

1 Answer

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.
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