# Combinatorial Probability Analysis

### Preface

Many problems in the theory of opportunity can be solved simply by
counting the many different ways in which a work can occur. The mathematical
theory of enumeration is known as combinatorial analysis.

### Basic Principles of Chopping

The following enumeration principles are very basic for all of our
work later. This principle simply says that if one experiment has the
possibility of an experimental outcome (outcome) and if another experiment has
the possibility of experimental results, then these two experiments have
probable experimental results.

*Suppose that two experiments will be carried out. Then if experiment 1 produces one of the probable results of the experiment and if, for each trial result 1, there is a possibility of experimental results from experiment 2, then the two experiments have the possibility of experimental results.*

EVIDENCE OF BASIC PRINCIPLES: This basic principle can be proven
by enumerating all possible outcomes of the two experiments as follows,

*(1, 1), (1, 2), ...,(1, n)*

*(2, 1), (2, 2), ...,(2, n)*

*(m, 1), (m, 2), ...,(m, n)*

In
this case we say that the results of the experiment are

*(i, j)*if from experiment 1 the results of the second experiment appear and from experiment 2 the*j-results*appear. Thus the set of all the results consists of m lines and each row contains n elements.
Solution:
By considering the selection of men as the results of trial 1, and choosing
women as the results of experiment 2, then from the basic principle we get
there are

*5 x 8 = 40*possible teams.
The
broader version of the basic principle says that if there are r experiments in
such a way that the first experiment has n1 probable results, and for each
possible experimental result of

*(i-1)*the first experiment is i,*i = 1,2, ...,*from the results of experiments from the first experiment, all of the*r*experiments have*n1n2 ... nr*possible experimental results.**Examples of Combinatorial Analysis Questions 1**

A
student committee consists of 3 freshmen (level 1), 4 sophomore (level 2), 5
junior (level 3), and 2 seniors (level IV). In this committee there is a
section consisting of 4 people each person must take from each level. How many
possible sections can be formed?

Solution:
From the general version of the basic principle, there are

*n1 x n2 x n3 x n4*= 3 x 4 x 5 x 2 = 120 possible sections that can be formed.**Examples of Combinatorial Analysis Questions 2**

A flat
vehicle number consists of 3 letters followed by 4 numbers. How many vehicle
number plates can be made?

Settlement:

It is
known that the number of letters from

*A-Z*is as many as 26 letters while the number of numbers from numbers*0-9*is as many as 10 numbers. Where will be arranged into a flat number vehicle with an arrangement of 3 letter slots and followed by 4 number slots. According to the general basic principle version 26 letters x 26 letters x 26 letters x 10 digits x 10 digits x 10 digits x 10 digits = 26 x 26 x 26 x 10 x 10 x 10 x 10 = 175 760 000 flat numbers.**Examples of Combinatorial Analysis Questions 3**

How
many functions can be formed if the area of definition consists of n points
and the function value must be 0 or 1?

Solution:
Suppose that the points are

*1, 2, ..., n.*Because*f(i)*must be 0 or 1 for every*i = 1, 2, ..., n*then there are 2n possible functions.**Examples of Combinatorial Analysis Questions 4**

In
this last example, how many vehicle number plates can there be repeated letters
or numbers?

Settlement:
Because there are conditions that there should be no letters or numbers
repeated So every 1 letter or number is taken then there is a process of
reducing the number of letters and numbers that will be used next, namely

*n x (n-1) x (n-2)*. So, in this example problem, the correct answer is 26 letters x 25 letters x 24 letters x 10 digits x 9 digits x 8 digits x 7 digits = 26 x 25 x 24 x 10 x 9 x 8 x 7 = 78 624 000 arrangement of flats number.
The
last problem is entered into the permutation calculation.

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