A person hosts a party of 13 guests and places 5 of them on one table and the remaining on the other table, the tables being round. In how many ways can he arrange the guests?

Let me break this down into simple steps that anyone can follow!

We have $13$ guests total. We're putting $5$ guests at one round table and $8$ guests at another round table. We need to find ALL possible ways to do this.

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First, we need to decide which $5$ guests (out of $13$) sit at the first table.

Number of ways to choose $5$ guests from $13$ = $\binom{13}{5}$

Why do we use combinations here? Because at this step, we only care about WHO sits at which table, not WHERE they sit at the table.

$\binom{13}{5} = \dfrac{13!}{5! \times 8!} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$

The remaining $8$ guests automatically go to the second table.

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Now we arrange the people around each circular table.

Key concept: For a round table with $n$ people, we fix one person's position (to avoid counting rotations as different arrangements) and arrange the remaining $(n-1)$ people. This gives us $(n-1)!$ arrangements.

First table ($5$ people): $(5-1)! = 4! = 24$ ways

Second table ($8$ people): $(8-1)! = 7! = 5040$ ways

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Since each choice is independent, we multiply:

Total arrangements = (Ways to choose groups) $\times$ (Ways to arrange table $1$) $\times$ (Ways to arrange table $2$)

Total = $\binom{13}{5} \times 4! \times 7!$

$Total = \dfrac{13!}{5! \times 8!} \times 4! \times 7!$

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Let me show you a shortcut:

$\dfrac{13!}{5! \times 8!} \times 4! \times 7! = \dfrac{13! \times 4! \times 7!}{5! \times 8!}$

Since $5! = 5 \times 4!$ and $8! = 8 \times 7!$:

$= \dfrac{13! \times 4! \times 7!}{(5 \times 4!) \times (8 \times 7!)} = \dfrac{13!}{5 \times 8} = \dfrac{13!}{40}$

Final Answer: $\dfrac{13!}{40}$