Question:
Let $K$ be the number of ways $16$ students can be arranged around a square table having $4$ seats on each side.
If $K = a(b!)$ with $b$ is maximum, find value of $a + b$.
My method:
I took $4$ members as a string, and I got $4$ strings and then I can arrange them in $3!$ ways. So total arrangements= $3!(4!\cdot4!\cdot4!\cdot4!)$
As I can arrange $4$ members in $4!$ ways and then fundamental rule of multiplication.
Best Answer
We know that had they been seated in an unnumbered circle, the number of arrangements $=\frac{16!}{16}$ because there are $16$ possible indistinguishable rotations.
For a square, there are $4$ possible indistinguishable rotations,
thus # of arrangements = $\frac{16!}{4} = 4\cdot15!\implies a+b = 19$