$\begin{array}{1 1}(A)\;\text{True}\\(B)\;\text{False}\end{array} $

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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