Counting Rules: Place Filling Rules, Permutations, Combinations

Counting Rules: Place Filling Rules, Permutations, Combinations – What is meant by the Enumeration Rule? On this occasion About the knowledge.co.id will discuss about the Enumeration Rule and the things that surround it. Let's look at the discussion together in the article below to better understand it.

Counting Rules: Place Filling Rules, Permutations, Combinations

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The enumeration rule is a counting rule to find out the number of certain events or objects that appear. It is called enumeration because the result is in the form of a whole number.

The enumeration rule (Counting Rules) is defined as a way or rule to calculate all the possibilities that can occur in a particular experiment. There are several methods in the enumeration rules including: the place filling rule method (Filling Slots), permutation method and combination method.

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Place Filling Rules

Problems:

Anton has 3 shirts which are white, red and blue and has 2 trousers which are black and brown. Determine the possibilities – the probability that Anton will wear a shirt and trousers!

Resolution:

There are 3 ways to determine the probability that Anton is wearing a shirt and trousers.

• • The set of ordered pairs

{(White, Black), (White, Brown), (Red, Black), (Red, Brown), (Blue, Black), (Blue, Brown)}

From the three ways above, it can be concluded that there are many ways that Anton wears shirts and pants
length = 6 ways = 3 × 2 = number of ways to wear a shirt × number of ways to wear pants
long.

• Multiplication Rule
  

If an event can occur in n successive steps where stage 1 can occur in q₁ way, stage 2 can occur in q₂ way, stage 3 can occur in q₃ and so on until the nth stage can occur in q_n way then the events can occur sequentially in q₁ × q₂ × q₃ × … × q_n different way.

Example :

In how many ways can 3 student council administrators, consisting of a chairman, secretary and treasurer, be chosen from 8 students?

Resolution:

There are 3 places to fill the positions of chairman, secretary and treasurer as follows:

Chief Secretary Treasurer

Of the 8 students, all of them are entitled to be elected as chairman so there are 8 ways to fill the chairperson's position. Because 1 person has become chairman, only 7 people are left who have the right to be elected as secretary, so there are 7 ways to fill the secretary position. Because 1 person has become chairperson and 1 person has become secretary, only 6 people are left who have the right to be elected as treasurer, so there are 6 ways to fill the treasurer.

8

7

6

Chief Secretary Treasurer

The number of ways to choose the 3 student council administrators is 8 × 7 × 6 = 336

• Addition Rules

Suppose an event can occur in n different (alien) ways where in the first way there are p₁ different possible outcomes, in the second way there are p₂ different possible outcomes, in the third way there are p₃ different possible outcomes and so on until the nth way there is p_n different possible outcomes, the total number of possible events in that event is p₁ +p₂ +p₃ + … + p.s_n different way.

Example :

Hendro is a SMK student. Hendro has three types of transportation from home to school, namely bicycles (mini bikes, mountain bikes), motorbikes (yamaha, honda, suzuki) and cars (sedans, deer, pick-ups). How many ways can Hendro get from home to school?

Resolution:

The only means of transportation used by Hendro from home to school is a bicycle or bicycle motorbike or car, there was no way he could use more than one vehicle at a time together. The number of ways Hendro can go from home to school is the number of ways to use a bicycle + the number of ways to use a motorcycle + the number of ways to use a car = 2 + 3 + 3 = 8 ways.

• Factorial Notation

Let n Î be the set of natural numbers. Notation n! (read: n factorial) is defined as the product of natural numbers sequentially from n to 1.

Written n! = n × (n – 1) × (n – 2) × … × 3 × 2 × 1.

Defined 1! = 1 and 0! = 1.

Example :

1. Determine the value of 5!.

Resolution:

5! = 5 × 4 × 3 × 2 × 1 = 120.

1. Determine the value of 2! + 3!.

Resolution:

2! + 3! = (2 × 1) + (3 × 2 × 1) = 2 × 6 = 12

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Permutation

Permutation is an arrangement that can be formed from a collection of objects that are taken in part or in whole by paying attention to the order. "Paying attention to the order" means that the arrangement AB and BA are considered to be different events. For example, in a class, 3 candidates have been selected to occupy the positions of chairman, secretary and treasurer. The three selected candidates are A, B and C. The possible composition of the management of the class is as follows:

There are 6 possible management arrangements.

Types of permutations:

• Permutations of n elements from n different elements

There are many ways to arrange n elements taken from n elements by paying attention to the order expressed by P(n, n) or nPn which is formulated as follows:

P(n, n) = n!

Example 1 :

Of the 4 candidates for the OSIS board, how many possible arrangements can be formed to determine the chairman, deputy chairman, treasurer and secretary at the same time?

solution:

The composition of the OSIS board candidates formed is P(4,4) = 4! = 1 x 2 x 3 x 4 = 24.

Example 2 :

Determine the arrangement of letters that can be formed from the word "LUANG" if the arrangement of letters consists of five different letters.

Resolution:

The possible arrangement of letters is P(5,5) = 5! = 1 x 2 x 3 x 4 x 5 = 120.

• Permutations of k elements from n different elements (k ≤ n)

There are many ways to arrange k elements taken from n elements by paying attention to being expressed as P(n, k) or nPk which is formulated as follows:

Example 1 :

Determine the number of possibilities in selecting the class president and vice president if there are 6 candidates.

Resolution:

Number of possibilities = P(6,2) = 30

Example 2 :

From the letters A, B, C, D, E, F, determine the arrangement of letters consisting of 3 different letters.

Resolution:

Number of letter arrangements = P(6,3) = 120

• Permutations with some of the same elements.

If of the available n elements there are n₁ same element, n₂ elements are the same and so on then the number of permutations is

Example :

Find the number of different letter arrangements in the word ACCOUNTANT

Resolution:

Number of letters (n) = 7, number of letters A = 2, number of letters N = 2

• Cyclic permutation

Observe the following picture! What do you think about this picture? Explain!
Cyclical permutation is a way to determine the arrangement of elements that are arranged cyclically or circularly by paying attention to the order. The number of cyclic permutations of n different elements is: P = (n – 1)!

Example :

In a meeting, there are 8 participants who will occupy 8 chairs around a round table. How many arrangements are possible?

Resolution:

Number of possible arrangements = (8 – 1)! = 7! = 5040.

• Repeated permutations

The number of permutations of r elements taken from the available n elements with each available element may be written repeatedly is P = n^r

Example :

How many arrangements of 3 letters are taken from the letters K, A, M, I and S if the available elements can be written repeatedly.

Resolution:

Number of arrangements = 5³ = 125.

Combination

Combination is an arrangement that can be formed from a collection of objects (each object is different) taken in part or in whole without regard to the order / randomly or randomly random. For example, if the refrigerator contains tape, pineapple and fro, then the way the ice seller puts the contents of the ice into the glass can be (tape, pineapple, and fro), (tape, fro and fro, pineapple), (pineapple, tape, fro and fro), (pineapple, fro and fro, tape), (pineapple, fro and fro, tape), (and fro, pineapple, tape) and (fro and fro, tape, pineapple). No matter how you put the contents of the ice into the glass, the result will be the same, namely combination ice containing the 3 types earlier. Combinations of r elements from the available n elements are formulated

Example 1 :

There are 12 basketball players who will compete. In the first minutes, 5 people will be deployed. How many possible ways can this happen?

Resolution:

The number of possible ways this could be C(n, r) = 792

Example 2 :

Three balls are drawn from a box containing 5 red balls, 3 white balls and 2 blue balls. Find the number of ways to draw three balls consisting of 2 red balls and 1 blue ball.

Answer :

There are 5 red balls available and 2 balls will be taken, there are many ways to get them

= C(5,2) = 10.

There are 2 blue balls available and 1 ball will be taken, there are many ways to collect them

= C(2,1) = 2.

The number of ways to draw three balls consisting of 2 red balls and 1 blue ball is 10 × 2 = 20.

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