# Counting using Equivalence Classes & Partitions : Squirrels in a square

• Oct 5th 2011, 12:22 PM
onemachine
Counting using Equivalence Classes & Partitions : Squirrels in a square
A) In how many ways can 36 different squirrels stand in a square? That is, four sides with nine squirrels on each side.

B) If there are 18 male squirrels and 18 female squirrels, in how many ways can these squirrels stand in a square if they must alternate in gender.

Both parts of this problem assume each squirrel is unique, just as if they were people.
• Oct 5th 2011, 12:34 PM
Plato
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
Quote:

Originally Posted by onemachine
A) In how many ways can 36 different squirrels stand in a square? That is, four sides with nine squirrels on each side.
B) If there are 18 male squirrels and 18 female squirrels, in how many ways can these squirrels stand in a square if they must alternate in gender.

These are just circular arrangements.
If N different objects is a circle, then there are $(N-1)!$ ways to do that.

If we have K of one kind and K of the another kind then there are $(K-1)!\cdot K!$ ways to arrange these so as to alternate kinds.
• Oct 5th 2011, 12:41 PM
onemachine
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
Quote:

Originally Posted by Plato
These are just circular arrangements.
If N different objects is a circle, then there are $(N-1)!$ ways to do that.

If we have K of one kind and K of the another kind then there are $(K-1)!\cdot K!$ ways to arrange these so as to alternate kinds.

Thanks for the response, but this is NOT a circle problem. This was made clear by my prof: The point of this problem is that it is NOT a circle problem and it is NOT a line problem. In a circle, if the squirrels move one position to the left or right, the circle remains the same circle. In a square, if the squirrels move one position, then we have a different square.
• Oct 5th 2011, 01:14 PM
Plato
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
What happens if the square is rotated: 90, 180, or 270?
• Oct 5th 2011, 02:31 PM
onemachine
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
Quote:

Originally Posted by Plato
If not, what happens if the square is rotated: 90, 180, or 270?

The way I understand it is that a square is a square if it has 4 unique sides. When the squirrels move one position to the left or right, there are 4 new unique sides created (and thus, a second, unique square). However, when they move positions 9 times, they have created the same square they started with. When they move to the 10th time, they are creating the same square they made after the first position. The squirrels can create equivalent squares while rotating throught the positions around the square. Thus, I believe the answer to part A is 36!/4, because the equivalence class of the original square has 4 elements, of which only one is a unique square.

I'm hoping someone can convince me that this is either correct or incorrect...
• Oct 5th 2011, 03:23 PM
Plato
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
Quote:

Originally Posted by onemachine
The way I understand it is that a square is a square if it has 4 unique sides. When the squirrels move one position to the left or right, there are 4 new unique sides created (and thus, a second, unique square). However, when they move positions 9 times, they have created the same square they started with. When they move to the 10th time, they are creating the same square they made after the first position. The squirrels can create equivalent squares while rotating throught the positions around the square. Thus, I believe the answer to part A is 36!/4, because the equivalence class of the original square has 4 elements, of which only one is a unique square.

I'm hoping someone can convince me that this is either correct or incorrect...

O.K. Would agree that there are $35!$ ways to make the first arrangement?
If so, can you argue that according to your understanding the answer is $\frac{35!}{4}~?$
• Oct 5th 2011, 03:37 PM
onemachine
Re: Counting using Equivalence Classes & Partitions : Squirrels in a square
Quote:

Originally Posted by Plato
O.K. Would agree that there are $35!$ ways to make the first arrangement?
If so, can you argue that according to your understanding the answer is $\frac{35!}{4}~?$

I do not see how there are 35! ways to make the initial arrangement. Since there are 36 squirrels and 36 positions on the square, there are 36! ways to arrange the squirrels.

If you draw a picture of the initial square, you can see that no single squirrel holds a position on two sides of the square (i.e. no single squirrel is on a corner). Thus, there are indeed 36 unique positions on the intital square.

Right? ;)