# Thread: Tricky Circular Arrangement Question (Permutations and Combinations)

1. ## Tricky Circular Arrangement Question (Permutations and Combinations)

So I have come across a question from Cambridge which is (unsuprisingly) difficult. If you can help me solve the problem (as I have no clue how to do it) that would be much appreciated. The question is as follows:

"There are two distinct round tables, each with five seats. In how many ways may a group of ten be seated?"

This is difficult as my class has only touched on circular arrangements today, and I am way too curious to not know how to solve it. The answer is supposed to be 145,152. If you can show me how to work it out that would be great. Thanks!

2. ## Re: Tricky Circular Arrangement Question (Permutations and Combinations)

$\binom{10}{5} \left(\frac{5!}{5}\right)^2 = 145152$

There are $\binom{10}{5}$ ways to pick 5 people from the group of 10. Then there are $\frac{5!}{5}$ ways to arrange each group around the table.

Normally we would have to consider the double counting case.

Suppose the 10 people are named A, B, C … J.

Using $\binom{10}{5}$ to pick five people means that picking $ABCDE$ is counted separately to picking $FGHIJ$, even though the same people are put in the same group each time.

However, in this case, since the tables are distinct, the order of the groups actually does matter, i.e. $ABCDE |FGHIJ$ is different to $FGHIJ |ABCDE$.

Possibly a less confusing way to think about it is:

$\frac{1}{2}\binom{10}{5} \left(\frac{5!}{5}\right)^2 \times 2!= 145152$

$\frac{1}{2}\binom{10}{5}$ gives the number of distinct ways in which 10 people can be divided into two groups.

Each group of 5 can be arranged around the table in $\frac{5!}{5}$ ways.

And there are $2!$ ways to assign each group of 5 to a table, since the tables are distinct.

3. ## Re: Tricky Circular Arrangement Question (Permutations and Combinations)

Originally Posted by fan96
$\binom{10}{5} \left(\frac{5!}{5}\right)^2 = 145152$

There are $\binom{10}{5}$ ways to pick 5 people from the group of 10. Then there are $\frac{5!}{5}$ ways to arrange each group around the table.

Normally we would have to consider the double counting case.

Suppose the 10 people are named A, B, C … J.

Using $\binom{10}{5}$ to pick five people means that picking $ABCDE$ is counted separately to picking $FGHIJ$, even though the same people are put in the same group each time.

However, in this case, since the tables are distinct, the order of the groups actually does matter, i.e. $ABCDE |FGHIJ$ is different to $FGHIJ |ABCDE$.

Possibly a less confusing way to think about it is:

$\frac{1}{2}\binom{10}{5} \left(\frac{5!}{5}\right)^2 \times 2!= 145152$

$\frac{1}{2}\binom{10}{5}$ gives the number of distinct ways in which 10 people can be divided into two groups.

Each group of 5 can be arranged around the table in $\frac{5!}{5}$ ways.

And there are $2!$ ways to assign each group of 5 to a table, since the tables are distinct.
Thank you very much! We haven't covered Combinations yet so that is most likely why I had trouble with it...

4. ## Re: Tricky Circular Arrangement Question (Permutations and Combinations)

Here’s another way to do it.

$\frac{^{10}P_5}{5} \times \frac{5!}{5} = 145152$

For the first table, use $^{10}P_5$ to obtain the number of ways you can arrange 5 people from a group of 10 (again, the concept of indistinguishable groups from before also applies here, but since the tables are distinct we do not need to divide by 2).

Because we are looking for a circular arrangement, divide by 5 (recalling the formula for circular permutations).

There are five people left, and there are $\frac{5!}{5}$ ways to arrange them on the remaining table.

5. ## Re: Tricky Circular Arrangement Question (Permutations and Combinations)

Originally Posted by fan96
Here’s another way to do it.

$\frac{^{10}P_5}{5} \times \frac{5!}{5} = 145152$

For the first table, use $^{10}P_5$ to obtain the number of ways you can arrange 5 people from a group of 10 (again, the concept of indistinguishable groups from before also applies here, but since the tables are distinct we do not need to divide by 2).

Because we are looking for a circular arrangement, divide by 5 (recalling the formula for circular permutations).

There are five people left, and there are $\frac{5!}{5}$ ways to arrange them on the remaining table.
Thank you so much!!!! You're a legend!

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