# Permutation: How to arrange 12 people around a table for 7?

I want to understand how to arrange $$12$$ people around a circular table with $$7$$ chairs. We don't care about the overflow, those people can go to another table.

I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $$11$$ possible options of people (since one person is already seated), the third chair has $$10$$ possible options, the fourth chair has $$9$$ possible options and so on until we get to the seventh chair which has $$6$$ possible options of people.

So I thought the way to solve is that this is a permutation problem $$1*11*10*9*8*7*6=332640=11P6=\frac{12P7}{12}$$. But my professor says the correct answer is $$\frac{12P7}{7}$$. I don't understand why we should divide $$12P7$$ by the number of chairs. Can someone explain this me?

• Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table. – JMoravitz Mar 10 at 20:07

I think it should be $$\binom{12}{7}6!$$
• Why multiply by $6!$? – Sam Mar 10 at 20:14
• @Sam Note that $\binom{12}{7}6!$ is equal to $\frac{_{12}P_7}{7}$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people. – JMoravitz Mar 10 at 20:30