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

Assalamualaikum Wr. Wb. Meet again with me, Denny Handayani on the MATLAB channel. In the previous video, we have learned the multiplication and multiplication rules. And this is the second part of the video. In this video, we will learn the rules for filling the place or filling slot. Okay, let's just discuss the material.

Okay, now we will learn the second part of the debugging procedure In this second part of the video we will learn the setting of the place or filling slot This rule is quite common for friends to use because there are quite a lot of problems or problems that we can solve using this rule

And this is the definition, for example there are n places available with k1 is the most way to fill the first place, k2 is the most way to fill the second place, and so on, until kn is the most way to fill the place to n. So the most way to fill the place is k1 times k2 times next to kn.

Yes, this is basically we use the rules of the year, friends Okay, so that you guys understand better, let's try to discuss some examples of the following questions First example, of the 8 OSIS members, 4 people will be selected as management teams to hold the head office, deputy head, secretary, and secretary How many management teams are possible? Well, we will use the rules of filling the place or filling slot

So here we will choose 4 people, so we provide 4 slots, 4 places

The first place is to hold the head office, then the second is the head of the office, the third place is for secretaries, and the fourth place is the embassy. There are 8 members of many management candidates, so we will fill in these slots, these places with a lot of possibilities. First we fill in the head section, how many ways do we choose a head? There are 8 candidates here.

8 managers who are all competent or can be the head of the head there are as many as 8 choices or 8 ways to choose a head now for the head of the head how many choices from 8 members

One member is definitely chosen as the chairman, right? So for the chairman, it's impossible for 8 more. It must be less than 1 because it has been chosen as the chairman. So here, many choices for the chairman are 7 ways. Then for secretaries, because of 8, 1 has been chosen as the chairman, 1 as the representative, then the rest ...

is 8 minus 2 means 6 ways to choose a secretary. The same goes for the governor, because there is no complete office here. From 8, 3 people have been elected as the chairman, deputy, and secretary. So for the governor, the rest is 5 people, or 5 ways to choose.

So how many possible management team arrangements? We use the management team arrangement rules, we multiply all the possibilities, friends So we multiply this by 8 times 7 times 6 times 5, there are as many as 1680 possible management team arrangements

Okay, easy huh? Now let's try the second example How many words can be made from the letters P, I, N, T, A, R? If the first condition, the first letter is a vowel Then the second condition, the first letter is a consonant Now let's solve these two problems

The first is the vocal, the vocal is A I U E O, yes, in the word P I N T A R, here the vocal is where? I and, okay, so here P I N T A R, there are 6 letters, so we provide 6 slots, 6 places, friends So how many ways do we choose the vocal letters in this section?

Because there are two, i and a, it means here we have two options, friends. There are two options from the word P, I, N, T, A, R, the vowel letters are two, namely i and a.

For the next letter, there is no requirement, so it's free, whether you want a consonant or a vowel, you can So from 6, it's already used 1, right? It's already taken 1, so here it's only 5 Then in this part, it's already used 2, so it's only 4, here it's only 3, 2, and 1, we just multiply it 2 x 5 x 4 x 3 x 2 x 1, the result is 240 So there are 240 words

For the second question, the first letter is a consonant, in the word PINTAR the consonants here are 1, 2, 3, 4, there are 4 consonants So for the first slot, there must be a consonant, then there are as many as 4 options Well for the next slot, there are no requirements here, so from 6

6 letters have been taken 1 then only 5 is left Well, the next one has been taken 2 then only 4 is left Next, subtract 3 again, here 2 and finally 1, we multiply this 4 5 4 3 2 1 we multiply the result is 480 So there are as many as 480 words that can be arranged with the first letter is a consonant

Now let's move on to the third example. The third example, from the numbers 1, 2, 3, 4, 5, and 6 will be formed a three-digit number. How many numbers can be formed if: The first, the numbers can be repeated.

The second, the numbers should not be repeated. The third, the odd number should not be repeated. The fourth, the number more than 300 with different digits, meaning it should not be repeated. And the last, the number less than 600 with different digits, meaning there should be no repeated digits. Okay, let's try to finish one by one. We finish the first part.

The condition is that the numbers can be repeated Here are 6 numbers that we can use And the number that is formed is 3 digits So we prepare 3 slots

For the first case, the number can be repeated, for example, the number that is arranged is 111, this is allowed, or 225, this is allowed, so there is a repeat, no problem. So for the first one, this is hundreds, the 20th slot, and the third slot is one.

So for hundreds, how many ways? Here are 6 numbers, right? So there are 6 choices for hundreds. Well for tens, because here it can be repeated, so the number that has been used in hundreds, we can use it again. So here it remains 6 friends. And for ones, because it can be repeated, it remains 6. So the number that has been used in tens and hundreds, we can use it again in ones. So many numbers are 3 digits.

The number that can be repeated is 6 x 6 x 6 = 216 numbers Now the second part, the numbers must not be repeated Here the same 3 digits, we prepare 3 slots for hundreds, tens and ones

Well for hundreds, there are 6 choices, 1, 2, 3, 4, 5 or 6, there are 6 choices. Well, because here it is not allowed to repeat, then the number or number that has been used in hundreds, it cannot be used again in tens. So for tens, not 6 choices anymore, but 5 choices. Because one of them has been used in hundreds. Well for one, from 6, 2 have been used in tens and hundreds, then there are only 4 choices left.

So, the number of numbers that can be formed that the numbers cannot be repeated is 6 x 5 x 4 = 120 numbers. Now, the third problem, if the number is odd and cannot be repeated.

The requirement of a odd or even number is seen from the units. So to form a three-digit number which is an odd number, then the units must be odd. From these six numbers, how many odd numbers are there? There are one, two, three.

So from these three slots we start from the units first. So that it forms a tick, then the units must tick. Here there are three ticks. So there are three ways we fill in the units.

The odd number is number 1, number 3, or number 5. There are three ways. From these 6, 6 numbers have been used 1 for 1. So for hundreds, the rest is 5. From 6, 2 have been used, so for tens, there are 4 more. So many odd numbers that do not repeat or each digit is different is 6 x 5 x 3 = 60 numbers. For section D, numbers more than 300.

With different digits, meaning it can't be repeated Well, more than 300, then we start from hundreds, friends So that more than 300 Then how many hundreds? 3 can not be hundreds? It's okay, right? 300 something, this is still more than 300, this is okay 4 of course, can be hundreds 400 something 5 can, 6 can Well, these are numbers that we can make hundreds How many numbers are there?

There are as many as 4 numbers, there are 4 choices for hundreds For tens, from 6 numbers, if 1 is used, then the tens are 5 For ones, from 6, if 2 is used, then for ones there are 4 choices So many numbers more than 300 with different digits are 4 x 5 x 4 = 80 And the last, numbers less than 600 with different digits

less than 600, then which one can be hundreds?

What can be hundreds, it's clear that this is 1, 2, 3, 4, or 5. 6 can't be hundreds. Because the number that is formed must be less than 600. So for hundreds there are as many as 5 choices. For tens also 5, yes, from 6 already used 1, right? For ones is 4, from 6 already used 2, right? So many numbers that are less than 600 with different digits are 5 times 5 times 4, as many as 100 numbers.

Okay, let's continue to the fourth example. The fourth example, how many numbers consist of three different numbers and are divided by 5 and can be arranged from the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

After divided by 5, then the possible units are 0 or 5. That's the characteristic of the number that has been divided by 5. Well, we divide this into two cases. The first if the units are 0, this is hundreds, this is tens, this is units. If the units are 0, then for the units there is only one choice, 0 only.

Now for the hundreds, here we have 10 numbers, already used 0 for the ones, then the choice for hundreds, only the rest, there are 9 choices

For the tens, from 10, 2 have been used, then the rest are 8 choices for the tens. So many numbers that are 0 are 9 times 8 times 1, 72 numbers. And the second case, the units are 5, if the units are 5.

Hundreds, tens, ones. For one, there is only one choice, namely five. Well, if the ones are five, now we will look for many choices for hundreds. Well, five of these have been used in the ones. The rest is nine. Nine numbers, but we can't put zero in hundreds, remember. Zero is impossible to put here. So for hundreds, only eight numbers, eight choices.

For tens, zero can be placed in tens, so there are 8 more. So the number of numbers that are one is 5 is 64. So in total, the three digits are different, which are divided by 5, there are 72, we add up to 64. There are 136 numbers.

The question is why the answer is not like this? Because maybe there are among friends why not just the units we fill in with 2? Because there are 2 possibilities 5 or 0 If the answer is like this, here hundreds are multiplied by 9, this makes it possible for 0 to be multiplied by hundreds So it can't be like this, it's better we just divide the case Okay, now we discuss the last example, the fifth example

Zaki will create an email address, for that need he needs a password or a password consisting of 8 characters. A good word for a password if combined between letters and numbers. Zaki will use his name on the first or last four characters in order. Then added with four different numbers from 0 to 9 randomly, for example Zaki1234 and so on.

Many of the words in the email that can be used by Zaky are, okay, this is about the 2019 national exam. Well, the possibility of the formatted password, remember the word Zaky is used in order, four initial or final characters in order. So it still forms the word Zaky, it must not be misused. So the possibility is that Zaky can be followed by four numbers.

Or it can also be 4 different numbers followed by the word Zaki, like this. Well here, it means we just have to focus on finding the possibility of these 4 numbers. There are many ways to arrange 4 different numbers from these 10 numbers. How many? It means we just have to multiply it. 10 * 9 * 8 * 7, which is 5040.

5040 for the first possibility and 5040 for the second possibility so the total possible password is 5040 we times 2 because there are two possible arrangements like this friends so the total is 10,080, says Sandy, okay, as a practice material, please try the following three questions,

That's all for this video, see you in the next video Assalamualaikum Wr Wb