Applying Counting Method - Speech or Presentation Example

Topic: Applying Counting Method Answer for Question one. The main difference between permutation and the combination is this. With permutation, every little detail matters. Dog, cat, and a goat are different from goat, dog, and cat. On the other hand, combination is much easier. The details do not matter, i.e. dog cat, and goats are the same as goat, dog, and cat. Lets say we have 5 school pupils: P1, P2, P3, P4, and P5 in a race. How many ways can they pick position one, two, and three? Let’s use permutation, since the order of the position they get matters.

In position one, we have 5 choices: P1, P2, P3, P4, and P5. (We make an assumption P1 takes position one).In position two, we have 4 choices: P2, P3, P4, and P5. (We make an assumption P2 takes position two.)In position three, we have 3 choices: P3, P4, and P5. (We make an assumption P3 takes position three.)We picked certain people to take the different positions. At first we had 5 choices, then 4, and then 3.The total number of options was 5*4*3=60. We had to pick three pupils out of the five.

To do this, we had to start with all the 5 options, then4, and, lastly, 3 when we ran out of the top three positions. The factorial for this is: 5! =5.4.3.2.1, but we only need 5*4*3 .So how do we get rid of 2*1? If we do 5!/2!, we get this:5!/2!=5.4.3.2.1/2.1=5.4.3we used 2, because this is what remained after picking the top three positions. This can further be expressed as: 5! / (5-3)! This means that we use the first three numbers of 5! On the other hand, combination is very simple, since the order does not matter.

Dog, cat, and a goat. Let’s figure out how many different ways these animals can be rearranged. We have 3 choices for the first animal, 2 for the second animal, and 1 for the last one. Therefore, we have 3*2*1 ways to rearrange the animals.2. Answer for Question two.Since there are 30 runners for the qualifier, the first run can be ran by all of them, the 2nd by 29, the 3rd by 28, the 4th by 27, and so on until the remaining individual is one in the track. Hence, the answer is 30 ways.3. Answer for Question three.

Number of ways = C (30, 8) = 30! / (22!*8!) = 5,852,925.4. Answer for the fourth question.We have to select 3 out of 30 and order is important so we have to use permutation.Number of ways = P (30, 3) = 30! /27! = 24,360.ReferencesGossett, E. (2009). Discrete mathematics with proof. (2nd ed.). Hoboken, N.J.: Wiley.Veena, G. R. (2005). Comprehensive basic mathematics. New Delhi: New Age International (P) Ltd., Publishers.