Kaidah Pencacahan 2 - Aturan Pengisian Tempat (Filling Slots) Matematika Wajib Kelas 12

Kaidah Pencacahan Bagian Kedua Understanding the Filling Slots Rule

Introduction to Filling Slots

• Deni Handayani introduces the topic of filling slots as part of the second video on counting principles, following previous lessons on addition and multiplication rules.

• The filling slots rule is defined: if there are n available slots, K1 represents the number of ways to fill the first slot, K2 for the second, and so forth until Kn for the nth slot.

Application of Filling Slots in Problem Solving

• The total number of ways to fill these slots is calculated by multiplying all possible choices: K1 × K2 × ... × Kn, emphasizing reliance on multiplication rules.

Example 1: Forming a Management Team

• A scenario is presented where 4 out of 8 OSIS members are selected for specific roles (Chairperson, Vice Chairperson, Secretary, Treasurer).

• For each role:

• Chairperson has 8 options (all members eligible).

• Vice Chairperson has 7 options (one member already chosen as Chair).

• Secretary has 6 options (two members already chosen).

• Treasurer has 5 options (three members already chosen).

Calculation of Total Combinations

• The total combinations for forming this management team are calculated as 8 times 7 times 6 times 5 = 840.

Example 2: Creating Words from Letters

• The next example involves creating words from letters in "pintar" with specific conditions:

• First condition requires the first letter to be a vowel.

• Second condition requires it to be a consonant.

Case A: First Letter as Vowel

• Identifying vowels in "pintar" gives two choices (i and a).

• Remaining letters allow flexibility without restrictions leading to calculations resulting in 240 possible words.

Case B: First Letter as Consonant

• Starting with four consonants allows multiple arrangements leading to 480 possible words when considering subsequent letter selections.

Example 3: Forming Three-Digit Numbers

• Various conditions are explored regarding forming three-digit numbers using digits from 1,2,3,4,5,6:

• Digits can repeat.

• Digits cannot repeat.

• Odd numbers only without repetition.

• Numbers greater than three hundred with unique digits.

• Numbers less than six hundred with unique digits.

Case A: Repeating Digits Allowed

• With six available digits and three slots filled freely leads to numerous combinations based on digit selection across hundreds, tens, and units places.

This structured approach provides clarity on how filling slots works within various contexts while applying mathematical principles effectively.

Counting Three-Digit Numbers with and without Repetition

Forming Three-Digit Numbers with Repetition

• The total number of three-digit numbers that can be formed with repetition allowed is calculated as 6 times 6 times 6 = 216.

Forming Three-Digit Numbers without Repetition

• When digits cannot repeat, the hundreds place has 6 options, the tens place has 5 options (one digit used), and the units place has 4 options left, resulting in 6 times 5 times 4 = 120.

Odd Three-Digit Numbers without Repetition

• To form an odd three-digit number, the unit's digit must be odd. There are three odd digits available (1, 3, and 5).

• After choosing an odd digit for the unit's place, there are still five choices left for hundreds and four for tens. Thus, the total is 6 times 5 times 3 = 90.

Counting Numbers Greater than Three Hundred

• For numbers greater than three hundred using different digits: valid hundreds are 3,4,5,6, giving us four choices.

• The tens place then has five remaining choices (after one is used), while the units have four left. This results in 4 times 5 times 4 = 80.

Counting Numbers Less than Six Hundred

• For numbers less than six hundred: valid hundreds are 1,2,3,4,5. This gives five options.

• The tens also have five choices remaining after one is used from six total digits. Units again have four left to choose from leading to a total of 5 times 5 times 4 =100.

Counting Divisible by Five Using Different Digits

Case One: Last Digit is Zero

• If a number ends in zero (to be divisible by five), it leaves nine options for hundreds and eight for tens since zero occupies units.

• Therefore, this case yields 9 times 8 times1 =72.

Case Two: Last Digit is Five

• If a number ends in five: only eight choices remain for hundreds (excluding zero).

• Tens can include zero but not repeat any previously chosen digit; thus yielding 8times8times1 =64.

Total Count of Valid Combinations

• Combining both cases gives a total of 72 +64 =136.

Creating Password Combinations

Structure of Password Requirements

• Zaki needs an eight-character password consisting of his name "Zaki" followed by four unique digits from 0–9.

Calculating Possible Combinations

• The arrangement allows either "Zaki" at the start or end followed by any combination of four different digits.

First Arrangement Calculation:

-[]( t872 s ) For "Zaki" first: Calculate combinations as 10times9times8times7 =5040.

Second Arrangement Calculation:

• Further calculations would follow similar logic if needed.

This structured approach provides clarity on how to calculate various combinations based on given constraints while ensuring all steps are documented chronologically with relevant timestamps linked directly to their respective sections.

Total Possible Passwords Calculation

Understanding the Calculation of Password Combinations

• The total number of possible passwords is calculated to be 5040, which represents one arrangement of characters.

• This figure is then doubled due to two potential arrangements, leading to a final total of 10,080 unique password combinations.

• The speaker encourages viewers to practice with three additional problems related to this concept as an exercise.

• The video concludes with a friendly farewell and an invitation for viewers to join in the next session.