A new way to make your teaching activity interesting

Well guys, I think this can be one of your references to find an interesting activity for your students in class. My friends and I made it a couple years ago for our senior high school students. I know that the issue I carried out in this project is a bit out dated. However, to make this activity success to engage your students' attention, you can carry out some most up to date issues today. 

Enjoy:)

 

My personal experience in teaching mathematics in the classroom

                                           

October 09^th, 2012 at 07.30 a.m., I still remember that on that day, I started my first day of teaching practice in the second grade of social science class at SMA N 44 Jakarta. Actually that was horrible at first but it was nice at the end. Yes, before starting to teach, as always I stood in the classroom and waited for all 38 students to find their seats. I did not know why but I thought I needed to do it in order to make sure that they were really ready to join with my class. I looked at every student in there. The fact is they were pretty quiet as they sat there and I loved it. I knew at that time, I got nervous for a moment. But, I forced myself to keep going and then focused on single student. Making eye contact with her and then moving to another around the room.
Without waiting any longer, I opened the class by reviewing the last material and then continued by giving some explanations and examples about the new material which was permutation. Yes, teaching permutation is not simply as you may think. Even though the topic is just talking about the arrangement, but still to make students understand the concept of permutation itself, there are some efforts that the teacher needs to do. It is not enough to give students the definition and the formula only, but also as a teacher, I know that I need to give some more explanations and show some examples and then relate them to practical applications through the various ways in order to make students easier build upon the concept of permutation. However permutation is a topic that provides students the foundation for lesson in probability. If students cannot master this topic then they will get some problems in the topic of probability.
Since there are many applications of permutations in everyday life, I thought that I needed to show the simple one first before showing the either complex. Yes, at that time, I began to pose the problem about someone who wants to create some various new codes that consist of 3 letters in it. The problem is started when the person needs to know the number of various codes that are formed if the person picks 3 letters only from five letters given which are A, B, C, D, and E if ABC is not the same as CBA or BCA. Again the purpose of posing this problem was to make students understand the concept of permutation, so by emphasizing on the words “if ABC is not the same as CBA or BCA”, perhaps they could find what permutation is.
I looked most students thought hardly of this problem. I saw some of them mentioned the probability of the arrangements as soon as I posed the problem while the others were busy to write down the probability of the arrangements before mentioned it out loud. I waited until 5 minutes long and then I let 2 students come in front of the class to write down their answers on the board. Shortly, they finished writing their answers. I looked at their answers and I found that their answers were slightly different at that time. I asked to the other students but they did not make sure which answer that was true.
As a teacher, I finally come up with an idea that could make students be able to find out the solution of the problem easily. I drew a box consists of 5 box in it and then I guided students to find out the probability of the amount letters can be filled on each box and then asked students to multiply all the options together to get the total number of outcomes.
The illustration:

So for the first letter, there are five choices. The choices are A, B, C, D, or E. For the second letter, there are only four choices since the same letter cannot be chosen twice. For the third letter, there are only three choices left to select from. Now multiply 5 x 4 x 3 to get the total number of permutations. So the answer is: when picking three letters from ABCDE, if ABC is not the same as CBA, there are sixty permutations.
Later I told to the students that this problem can also be done quickly by using the formula of permutation that is nPr where the n stands for the number of items total, in this case five (ABCDE) and the r is the number that you want to pick, in this case three. So using 5P3 will also get sixty. Yes, either using the manual method or using the formula, we will get the same answer.
Since this topic is necessary, I just thought to be more likely posing some examples of permutation and then its answer through some demonstration in front of students. Be logical and simply figure out what the options are for each piece of the arrangement until they get the answer, yes, that is the point that I want to emphasize in this topic.
However, if I have a chance to teach the same topic again, I would not change any of my examples. I will still use the example above and maybe for the addition I will give the others various problems that are more difficult and exactly can deepen students’ understanding toward this topic.   
In summary, I just want to say that when you know why these things work, you have power, knowledge, and an edge over this test. When you just learn formula, you have the potential to make a really big mistake.