1. In how many ways can you arrange a round table for three men and three women if you want males to be sitting next to females?

1. Men are only allowed to sit next to women. Or,
2. Men just need to sit next to one woman (at least).

I will consider case a) with a brief mention on case b). We have three men and three women, which I shall separate into m1, m2, m3, f1, f2, f3 respectively. Around a table the arrangements will resemble:

Now we can unfold this into a sequence such as m1, f1, m2, f3, m3, f2, m1, f1, … This sequence repeats over and over again because it is cyclic (hence the round table). Now how many different possibilities are there for this table arrangement. Well let me first look at it this way. We have six positions.

There are six possibilities to choose position 1 (m1, m2, m3, f1, f2, f3). Now for positions two we have three possibilities of the opposite sex to choose from.

Now how many for position three? Well we it must be the same sex as position one and one of them is already used up. So there are (3-1) = 2 ways to choose position three. Like wise for position four (having the same sex as position two).

Now for positions five and six there is only one choice here for both of them, this has been forced by the other positions (the pigeon hole principle). So the total number of possible arrangements are: 6 x 3 x 2 x 2 x 1 x 1 = 72 different ways (by the fundamental principle of counting). Now a special note should be made that since we are on a round table there is the possibility of repetition of sequences. This happens in this problem. These 72 cases reduce down to twelve cases which are non-repetitious. One can see that these cases do not repeat previous cases by extending the sequence out in both directions (in other words repeating it) and attempting to match up like terms in a corresponding sequence. Each sequence differs by at least one term (or person in this case). The twelve sequences are as follows:

Now by looking at the sequences one can see how they were formulated. I used a reducing forcing sequence, which varies one position while holding another fixed, and then changes these roles. Any doubles are easily spotted by picking one element and then proceeding down the sequence. For example these two are the same sequence:

m1, f1, m2, f2, m3, f3 and f1, m2, f2, m3, f3, m1

Now for a brief mention of part b) we could have something like the following,

Males and females can be next to one another but each male must be next to at least one female. Well how many can we have? We can have six possibilities for the first position. Now we can have five possibilities for the second position. The third position is must be a female. But the number of possibilities to choose a female from depend on if we have chosen a female in the second position (or not).

Similarly for the fourth position, and so on to the last position. Now if we picked a male in our first position and a male in the second position then a female must go in the last slot. Remember we are on a round table. As you can see the problem interpreted this way gets very hairy. A possible method of solution could incorporate a spanning tree to clearly list the possibilities-arrangements.

1. Is it possible to arrange n x n objects each one having one of n different shapes and having one of n different colors in a square formation, such that each row and column contains exactly one object of each shape and one of each color?

Do this for n={2, 3, 4, 5}.

N = 2

This case is impossible. We only four objects we will start by filling in the first row.

When we come to filling in the second row, we can easily pick a different shape but once it comes to the color we run into a problem. We cannot pick color 1 (that is taken care of already by the 1 x 1 entry. We cannot pick color 2 because then that would make this entry (2 x 1) the same as the (1 x 2) entry. And we are out of colors and shapes to choose from…ergo this is impossible. Now any other formulation, which is impossible, is accomplished by the same method.

Let’s try the

N = 3

We have it clearly! Notice how I did this. I started out randomly picking a few and then the rest were forced by the pigeon-hole principle. The other cases follow:

N = 4

N = 5

3. Suppose points are arranged in a plane so that no three of them are co-linear

Find the number of segments that can be drawn using the given number of points for end points.

The following table gives the results of a number of points up to ten and the number of edges between them. As we add on a new point we can connect this new point to each and every other point that is already present. This will give rise to a new edge for each point already present.

Number of points		Number of edges between them
0		0
1		0
2		1
3		3
4		6
5		10
6		15
7		21
8		28
9		36
10		45

So if we have two points we have 1 edge. When adding another point to give us three points we can connect this new point to the two points already present which will give rise to two more edges plus the one already there (so we have three total). Now adding another point we have the three edges already there plus the new edges formed by connecting this fourth point to the three points already present. This gives rise to 3 new edges. 3 + 3 = 6. In this way we can see that the number of edges will be as follows:

E = [n (n – 1)] / 2

* The number of edges for 8 vertices is listed in the table above.

To see this, look at the below diagram:



4. Describe what you consider to be the role of the teacher in helping students to learn how to solve problems, like the ones discussed.

In working the above problems it seems that the problems themselves were of an exploratory nature. The students were to get into groups and work through the problems, experimenting and testing as they saw fit. The teacher took the role of facilitator and consultant, smoothing out any unforeseen problems that may have arisen through the working. It is this role that I believe the teacher should take when working similar problems as the above. Acting as a facilitator to group discussion, and consultant of questions, allows for the students to interact among themselves and construct their own methodology when approaching similar problems in the future.

5. Reaction to a situation where students give multiple answers to a counting problem. How should the teacher react?

I think in this particular situation, the teacher should allow the students further justification of the methods used to construct their answer. Doing this in an orderly fashion. Questioning in a format so the students can spot out flaws in any argument (and thus their logic). Again the teacher should not point out the flaws specifically, but question in a manner which leads to the "correct" answer. In this situation the teacher should act as devils advocate, but all the while goading the students in the desired direction.