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Diego 12A4 (10) Compulsory Maths Portfolio
Topic: Kaidah Pencacahan (Probability, Permuatations, And Combinations)

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Published by diegoimmanuel88, 2021-11-11 02:01:13

Portfolio Diego XII MIPA 4 (10)

Diego 12A4 (10) Compulsory Maths Portfolio
Topic: Kaidah Pencacahan (Probability, Permuatations, And Combinations)

Keywords: Maths Probability Permuatiation Combination

Compulsory Maths

Probability Permutation and Combinations
, ,

portfolio kaidah Pencacahan

by Diego Ronaldo Immanuel Tambunan 12×-4/10

Mind Mapping K

Pelu

kejadian Majemuk

kejadian Saling lepas kejadian Saling bebas kesadian bersyarat

Tujuan Pembelagaran

Peserta didik dapat menjelaskan aturan perkalian dan penjumlahan, menganalisis aturan perkalian dan penjumlahan melalui masalah kontekst

Sumter : Buku matematika wajib Revisi 2018 Hal 85 Basic competencies
3.3 Menganalisis aturan pencacahan (aturan p
4.3 Menyelesaikan masalah kontekstual yang b

Portfolio List
1). Identitas diri +foto+logo 61 ✓
2) Peta konsep /mind mapping kaidah penc
3) Deskripsi gambar dan Tabel/diagram pem
4) Analisis perbedaan masalah aturan perk
5) Latihan soal Mandiri aturan penjumlaha
6) Penugasan permutasi ✓
7) Analisis situasi masalah permutasi dan
8) Latihan soal Mandiri kombinasi ✓
9) Screenshoot/foto penilaian keterampila
10) Catatan rangkuman materi ✓
11) Penilaian diri ✓

Kaidan Pencacahan

uang

Aturanpencacahan

- turan Penjumlahan Aturan Perkalian Permutasi kombincisl

Permutasi semua unsur benda Permutasidengan unsur Sama Permutasi siklik

tual, serta mampu menyelesaikan masalah kontekstual yang berkaitan dengan aturan perkalian dan penjumlahan

penjumlahan, aturan perkalian, permutasi, dan kombinasi) melalui masalah.
berkaitan dengan kaidah pencacahan (aturan penjumlahan, aturan perkalian, permutasi, dan kombinasi) kontekstual.

cacahan ✓
mecahan ✓
kalian filling slot ✓
an dan aturan perkalian ✓
kombinasi v
an yang dilaksanakan pada saat peserta didik mengerjakan soal dalam KBM ✓

Deskripsi G-ambar & Tabet / Diagram Pemecahan

S O AL Table

Faisal membeli 5 baju warna berbeda dan 3 celana panjang yang warnanya berbeda. A BC D E
Berapa banyak cara Faisal dapat menggunakan setelan baju dan kemeja tersebut?
1 At B1 c) DI El
ABC

2 A2 132 C2 D2 E- 2

3A 133 c. 3 ☐ 3 E3

DE Analysis

From the example above we can conclude that the formula is :

Banyak denis Basu ✗ Banyak denis Celano

Visual Proof

I 23
I 23

conclusion = 15 Cara kemesa = 5 variasi

5 ×3 Celona = 3 variasi

45 6 > 89

10 11 12 13 14 IS

Analisis Perbedaan Masalah Aturan Perkalian Filling Slot

4A

Terdapat angka-angka 7,5,6,3 Akan disusun bilangan terdiri dari 3 angka (boleh berulang). Total Angkor Yang dpt terbentuk 64
Tulislah angka berapa saja yang dapat kamu susun?
Ada berapa banyak angka yang terbentuk? Ratusan Puluhan Saturn
4 4 4
Angkor yg dpt dibentuk

333 335 336 337 353 355 356 357 363 365 366 367 373 375 376 377
555 553 556 557 533 535 536 537 563 565 566 567 573 575 576 577
666 663 665 667 633 635 636 637 653 655 656 657 673 675 676 677
777 773 775 776 733 735 736 737 753 755 756 757 763 765 766 767

4 ×4× 4 = 64 angka

Terdapat angka-angka 7,5,6,3 Akan disusun bilangan terdiri dari 3 angka berbeda (tidak boleh berulang). Total angka 49 dpt dibentuk 24 anglia
Tulislah angka berapa saja yang dapat kamu susun?
Ada berapa banyak angka yang terbentuk? Puluhan Ribnan Ratusan
4
Angkor 49 dpt dibentuk 3 2

356 357 365 367 375 376 4 ✗ 3 × 2 = 24 angka
536 537 563 567 573 576
635 637 653 657 673 675
735 736 753 756 763 765

* The amount decreases when you are not able to repeat a number

4B

Terdapat angka-angka 7,5,6, 3 Akan disusun bilangan ganjil terdiri dari 3 angka (boleh berulang).

Tulislah angka berapa saja yang dapat kamu susun? -

Ada berapa banyak angka yang terbentuk? Total Numbers made : 48

Numbers I can make Hundreds Tens units because to get odd numbers you can 'x
333 335 337 353 355 357 363 365 367 373 375 377 4 3
555 553 557 533 535 537 563 565 567 573 575 577 4 end with an even (3/5,7) not (6)

663 665 667 633 635 637 653 655 657 673 675 677 4× 4 ×3 = 48 numbers
777 773 775 733 735 737 753 755 757 763 765 767

Terdapat angka-angka 7,5,6, 3 Akan disusun bilangan genap terdiri dari 3 angka berbeda (tidak boleh berulang). Total numbers I 6

Tulislah angka berapa saja yang dapat kamu susun? - can make :

Ada berapa banyak angka yang terbentuk? It can't be 6 cause Hundreds Tens Units
←3 I→
Numbers I can make 6 Is used at the end 2 only 6 bc it is even

356 376 3 ✗ 2✗ I = 6 angka
536 576
736 756

4C Total numbers I can make : 75

Terdapat angka-angka 7,5,6, 3, 2 Akan disusun bilangan lebih dari 500 terdiri dari 3 angka (boleh berulang). HP S
Tulislah angka berapa saja yang dapat kamu susun?
Ada berapa banyak angka yang terbentuk? 35 5 → can be repeated

Numbers I can make t
5,6 , > → 500 >
555 552 553 556 557 522 523 525 526 527 532 533 535 536 537 562 563 565 566 567 572 573 575 576 577
662 663 665 666 667 622 623 625 626 627 632 633 635 636 637 652 653 655 656 657 672 673 675 676 677 3 × 5 × 5 = 75 angka
772 777 773 775 776 722 723 725 726 727 732 733 735 736 737 752 753 755 756 757 762 763 765 766 767

Terdapat angka-angka 7,5,6, 3,2 Akan disusun bilangan lebih dari 550 terdiri dari 3 angka (boleh berulang). Lebih dari 550 = 65
Tulislah angka berapa saja yang dapat kamu susun?
Ada berapa banyak angka yang terbentuk?. 1 H P5 More than 500
.
Numbers I can make 35 5 3 ×5 ×5 = 75 angka
I = 10 numbers
555 552 553 556 557 562 563 565 566 567 572 573 575 576 577 5,6, > less than 550
662 663 665 666 667 622 623 625 626 627 632 633 635 636 637 652 653 655 656 657 672 673 675 676 677
772 777 773 775 776 722 723 725 726 727 732 733 735 736 737 752 753 755 756 757 762 763 765 766 767 2 HP S
.
diatas 550 ltanpapuluhan 223 di 627 ) I ✗2 ×5
1 25
HT u HT U
33 Y ¥3

567567 5 = 45 2 2 5 = 20 451-20=65

67 23

Latihan Soul Mandiri Aturan Penjumlahan & Perkalian

Latihan Soul mandiri

I. Jika Seseorang berangkat dr Kota A menuju Kota E , berapa banyak altenatif galan Yang dipilih ?

1. A →B →c }→ E 121-9 = 21 Cara
3 ✗2
✗2 = 12
2. A →D →E
→B ✗I =9
3
✗3

2. -

-

Total :9numbersµ@9@ggg

① 5007 5,6/7/8,9 5 Probability

Angkor 1 =

② 5 can't be used so the 2nd digit has 8 probability

③ we used 2 I can't be the same so 3rd digit 7 probability

calculation 7 = 280 numbers

5×8 ×

Penilaiandiri Pertemuan I











Penugasan Permutasi

Suatu keluarga yang terdiri dari 6 orang duduk melingkar pada meja makan. Jika ayah dan ibu selalu duduk berdampingan, maka banyak cara posisi duduk melingkar anggota keluarga tersebut

adalah… . I

pad can't be changed

•••o• Mom n= 4 1- I= 5

•••• t b

other chairs mom & dad -_ always together

1st GO ooo 4th child Round Table formula ( 1)n - ! ✗ 2!
( n - 1) ! ✗ 2 !
child BO (5-1) ! ✗ ( 2. 1) f

BO 3rd child mom 1- dad
2ndchild
4! 2.

4. 3. 2. 1.2

24 . 2 = 48 ways

Dengan berapa cara 5 anak laki-laki dan 3 anak perempuan dapat disusun pada suatu lingkaran jika anak perempuan selalu berdekatan (berkumpul)
-

•• always together counted as I
Bo n=5 +1 =6

••• •••oB girl Round Table formula ( 1)n - ! ✗ 3!

•• ( n - 1) ! ✗ 3 ! t
008
3 girls

••o• (6-1) ! 13.2.1 )

girl 51 6

. .

oooo ( 5. 4. 3. 2- 1) . 6

girl 20.6 6
.

120.6 = 720 Ways

Dalam berapa cara, 6 buku pelajaran berbeda dapat disusun pada sebuah rak buku? → Permutation be order doesn't matter

n= amount of books = 6
r=
Places available =6

Formula

" Pr =n!_

( n - r) !

,g6÷GPO = = 5-4-1,2-2 = = 120

,,

Dari sejumlah siswa yang terdiri dari 3 siswa kelas X, 4 siswa kelas XI dan 5 siswa kelas XII, akan dipilih pengurus OSIS yang terdiri dari ketua, wakil ketua dan sekretaris. Ketua harus selalu berada
dari kelas yang lebih tinggi dari wakil ketua dan sekretaris. Banyak cara untuk memilih pengurus OSIS adalah… .

10th grade = 3 Student Council Total of ways =
11th grade = 4
12th grade =s 1 Chairman ( > VCIS ) 24 t 210 = 234 ways
.

2- Vice chairman
3. Secretary

2 options : 12th grade chairman V5 11th grade chairman

11 grade probabities 12th grade probabities

chairman = 4 ( 11th grade ) chairman = 5 ( 12 grade)
( 10th grade)
Vice =3 vice chairman _- 7( 10th or 11th) → -2 be there r only 2 spots)
Secretary = 6(
secretary = 2 ( other vnchosen 10th grade 5 ✗ >✗ 6= )10th or 11th
4 × 3 ×2
= 24 210

Untuk keamanan di suatu bank, nasabah diminta membuat kata sandi dari susunan 4 huruf dati kata “aman” dan diikuti 2 angka yang tidak boleh sama. (contoh: mana71, naam54, dsb.).
Banyaknya kata sandi yang dibuat adalah… .

* Find the option for "Aman "

AMA N by 2

-
double so divided

41 " = 12
=

* Now The numbers 2

Total numbers 10 ( 0, I 2,314 , 5,6 7/8,9 )
, ,

permutation because order doesn't matter ( 2,0 ) diff (0/2)

N = options = 10

r = chosen = 2

÷._rFormula " Pr = ,!

l0÷/" Pz
= = go ways
=

(10-2) !

* Total 12 = 1080 ways
90

.

Penilaian Diri Pertemuan ke 2











Combination

The combination of r elements taken from the available n elements ( each element Is different)

r elements disregards the order ( rsn)

ncr = n!

( ) !r !
n -r

{= 6! = 6! 6. 5.4T 30 = 15
4 2.1
4! 2 ! ==
4 ! (6-4) !
14.1.2 !

"C > = 11 ! 11 ! ? É+== 330
11 10 g. 8 # 11 10 9-
= .. = ..
.

7 ! (11-7) ! 7 ! 4 ! H . 4! 14.7.21 I
.
. 1

12 = 12 ! 12 ! 12.11.10 # 24 220

[3 = 3! I 9!) = . +2 11 10 =
..

=

3 ! (12-3) ! 3 ! 9T X E. I
. .
,

72 = 28 nilain ?

= n ! = 28

2 ! ( n - 2) !

= n In - 1) ( NH) ! = 28
.
2. 1 ( nfa) !

n2 - n = 56

m2 56- n - = 0

:-&

In - 8) Vlnt > )

n=8 V n= -7

n >r

n 21 nilai n

Cn =
-2

n! = 21

( n -2) ! . )(n - In -21 21

n.cn -11 ! .tn/-2) ! =

nf( ) ! ( )n n- -12

( 1)n
.
!n - = 21

2

n2 - n = 42

n2 - n - 42 = 0

f- 7
6

n -7 V n +6

n=⑦ n = -6

n[ 4 = 35 , n2

n! = n ( n - 1) ( n - 2) (n -3) ( Nfl) = 35 n= 7

4 ! / n - 4) ! 4. 3. 2. t.cn/-4 ) 35.24 n2= 49
.

( ) ( 3)m2 - n
( n -2 ) n- =

n4 - 6h3 + 11h2 - 6h - 840=0

( nt4) ( n -7) ( NZ - 3h 1- 30 ) = 0

TM TM

" ( 3.SC %¥!10 ! 10.9-3.8-4 5 1 = 4-0
,
" Cy = 3%-37 - = -15.7 13 91
. .

15 ! F. 2- I
.
4 ! (15-4) É15 13 #
..
.

#= 4-3 ZI
10 . g. 8.7T 5 .
.

.

.

3 ! .tt 4T

15.14.13 . 12 #
.

4! #
.

3 ^ " Cz = 7 " Cz

(3 ( NH) ! => n!_
3 ! Intl - 3) 2 ! ( n - 2) !

✗ .tn/-2)(ntllnfnl) X( nM.)2!.(=n>/-2#2-.C)n-)2!

( n 1- 1) = 7

n = 7-I

=6

Anilisislah kedua Masalah Berikut

Terdapat 4 peserta didik Amir, Budi, Cahya, dan Doni. Akan dipilih 3 orang untuk bermain peran dongeng sebagai kancil, semut, dan cicak.

Apakah sama Jika terpilih Amir sebagai kancil, Budi sebagai semut, dan Cahya sebagai cicak dengan Terpilihnya Amir sebagai semut, Budi sebagai kancil,
dan Cahya sebagai cicak?

Berapa banyak cara mereka memainkan peran?

From this problem, if Amir was chosen as a mouse deer, Budi as an ant, and Cahya as a lizard, it will be different. If Amir was chosen as a mouse deer, Budi as a
ant, and Cahya asa lizard.
When Budi plays as an ant it will not be the same as Amir playing an Ant

Options for the Play using permutation

4133 = 4 !

(4-3)

4 3 2= . I
..

1

= 24 Ways

Terdapat 4 peserta didik Amir, Budi, Cahya, dan Doni. Akan dipilih 3 orang untuk mengikuti cerdas cermat.

Apakah sama Jika terpilih Amir, Budi, dan Cahya
dengan Terpilihnya Budi, Amir, dan Cahya

Berapa banyak cara memilih mereka menjadi tim cerdas cermat?

Based on the question, if Amir, Budi, and Cahya are elected, it will be the same as if Budi, Amir, and Cahya chosen.
So, ABC will equal to BAC

Choosing The Quiz Team using combinations

4C = 4! = = 4 ways
,
3 ! (4-3) !

Combination Practice Questions

Ada 4 orang anak, akan dipilih 3 orang untuk menjadi pengurus kelas sebagai ketua, sekretaris dan bendahara. Berapa banyak cara memilih 3 orang tersebut?

Permutasi

4ps = 4 !

(4-3) !

4 ! 2=
43 I = 24 Way
= .. .

1! 1

Ada 4 orang anak, akan dipilih 3 orang untuk mengikuti seminar. Berapa banyak cara memilih 3 orang tersebut?

Using combination

4cg -= 4 ! = = 4 ways

(4-3) ! 3 !.

Jika terdapat 10 soal dalam ujian. Peserta didik diminta mengerjakan 8 soal dari 10 soal tersebut. Berapa banyak cara peserta didik dapat mengerjakan soal tersebut?

10cg = to ! 10 ! to 9 # = 45 Ways
..
( 10.8 ) ! 8 ! = = 2- . 8T

21.8 !

Jika terdapat 10 soal dalam ujian. Peserta didik diminta mengerjakan 8 soal dari 10 soal tersebut. Jika nomor genap wajib dikerjakan Berapa banyak cara peserta didik dapat mengerjakan soal
tersebut?

'5cg = 5! even numbers : 2 i 4 , 6 , 8- 10

(5-3) ! 3 ! ? #5 4- = 10 ways
.
= 5!
= I 34
2! 3!

Pada sebuah lingkaran, terdapat 8 titik yang berbeda. Dengan menggunakan kedelapan titik tersebut, banyaknya tali busur yang dapat di buat adalah…

To make 1 bowstring from 8 different points we need 2 points.
Using Combination

°C 2 = 8 ! 8! = 8 7. 6 ! = 28 Way
.
=

(8-2) ! 2 ! 6 ! 2! 6! 2
.

Diketahui terdapat 4 titik sembarang (tidak ada 3 titik yang tak segaris), yaitu titik A, B, C dan D. berapa banyak garis yang dapat di bentuk dari 4 titik tersebut?

4C = 4! 4! = #.3.Z!_ = 6 ways
2
(4-2) !2 ! = -2 -2T
,
2.2 !

Dalam sebuah pertemuan terdiri dari 3 orang, dimana setiap orang berjabat tangan satu kali dengan setiap orang lainnya dalam pertemuan tersebut. Berapa banyaknya jabat tangan yang terjadi??

>( 2 = 3! 3! 3. 2T 3 ways

= ==

(3-2) ! 2 ! 1! 2! 2T

Dalam sebuah pertemuan terdiri dari 10 orang, dimana setiap orang berjabat tangan satu kali dengan setiap orang lainnya dalam pertemuan tersebut. Berapa banyaknya jabat tangan yang terjadi?

10C 2 = 10 ! 10 ! 10.9.8 ! = 45 Ways
(10-2) ! 2 !
= =

0! 2 ! 8! ✗ 2

Dalam sebuah pertemuan sejumlah orang, dimana setiap orang berjabat tangan satu kali dengan setiap orang lainnya dalam pertemuan tersebut. Jika terjadi 15 jabat tangan, berapakah jumlah
orang yang adadalam pertemuan tersebut?

Is = ×! So there are 6 people in the meeting

1×-2) !2 !

15 = ✗ ( x - 1) ( x -2) !

( x - 2) ! 2 !

15 = ✗ 2 ✗-

2

0= ✗ 2- ✗ - 30

( ✗ + s ) ( x - 6)

✗ = -5 V ✗= 6
TM

If there are 10 questions in the exam. Students are asked to work on 8 questions out of 10 questions. If all odd numbers are mandatory, in how many ways can students work on the problem?

odd numbers = 1 , 3,5 , 7, 9

leaves us 3 questions to be done ( 8-5 = 3 question)

5cg = 5! 5! ¥5 # 10 ways

== 12 3T =
.
( 5. 3) ! 3 ! 2! 3!

In an exam there are 10 questions from number 1 to number 10. Examinees are required to work on questions number 1, 3, and 5 and only work on 8 of the 10 questions available.
How many ways do examinees choose the questions to work on?

Questions that must be done = 113,5
Leaves us 5
→ quest that has to be done

questions opt to be done ( 8-3=5 )questions

Available questions 10 3 7= =
-

to I t

avail compass available options

>Cs = 7! 3
#7 -6
= . = 21 ways to work on
.

5! . 2! 5T . 2- I

If there are 5 questions in the exam. Students are asked to work on 3 questions out of 5 questions. If all number 1 must be done, in how many ways can students work on the problem?

must be done = I

3 quest must be done ( 3- I = 2) , so 2 question opt that can be chosen

to

comp

out of the 5 we have to do 1 (5-1--4) so 4 available questions
,

4C , = 4! = 4.3.tl#--12- = 6 Ways

2 ! (4-2) ! It 2 ! 2

In a fish pond, there are 5 koi fish and 4 mujair fish. Pa Ali will fish 1 fish from the pond. How many ways can Pa Ali get 1 Koi fish?

* bc we r looking for Koi , Just count Koi

5C , = 5! = 5 4T 5 ways
.=
(5-1) ! 1 !
. 4T l !
-

In a fish pond, there are 5 Koi fish and 4 Mujair fish. Pa Ali will fish 2 fish from the pond. How many ways can Pa Ali get 1 Koi fish and 1 tilapia fish?

5C 4C specific/y asked
,
,.

cs-F.TT 4 ! 50 times
-
! (4-1) ! . 1 !

5.LT! . 4 3T 20 Ways
. =
4T
. 3T

In a fish pond, there are 5 koi fish and 4 mujair fish. Pa Ali will fish 2 fish from the pond. How many ways can Pa Ali get the same fish?

add

5 (z t 4 Cz

5! 4 !
+
( 5. 2) ! -2 ! (4-2) ! 21
.

5.tt?3!-t4-?3.2!3t
. 2- 2T 2-

10 t 6 = 16 ways

In a fish pond, there are 5 koi fish and 4 mujair fish. Pa Ali will fish 3 fish from the pond. How many ways can Pa Ali get at least 1 koi fish?

3 fish options

1 Koi & 2 tilapia 2 Koi & 1 tilapia 3 Koi

5C , , " Cz ' Cz I 4C 5cg
,

5! 4! 51 . 4 5!
.
(5-1) ! 1 ! , (4-1) ! 1 ! (5-3) ! -3 !
Is -211.2 !
(4-2) ! -2 !

,-.i¥5.4.H.si5- 4.3! 4 3!

.

,

3 ! -2 3! 2 ! 3- !

5 6= 30 ways 10 4 = 40 ways 5.gg#2-- 10 ways
.
.

Total = 301-40 + 10 = 80 ways

A box contains 5 green balls and 4 red balls. Two balls will be drawn at random.
In how many ways can 1 green ball and 1 red ball be selected?

5C 4C ,.
,
5! 4!

4! 1! '

3! - l!

5. 4T = 20 Ways

¥Z.

4T

A box contains 5 green balls and 4 red balls. 3 balls will be drawn at random.
In how many ways are there to choose 2 green balls and 1 red ball?

4C5C . ,
z

5 ! 4!
.

31.2 ! 3! l!
-

5.251.3-1 4. 3-!
.
. #

3T . I

40 ways

A box contains 5 green balls and 4 red balls. 3 balls will be drawn at random.
In how many ways can all three be red?

" Cz " ways

4!

(4-3) ! 3 !

Y;t¥- =

A box contains 5 green balls and 4 red balls. 3 balls will be drawn at random. How many ways to choose at least 2 reds?

The Options Total = 301-4=34 ways

2 reds & 1 green 3 reds
4C " Cz
5C .
, z

5! . = s.IT?-.-4zj?j?!- = 5.6 4z?41 = = 4 ways
4 ! 1! 30 ways .
2 ! -2 ! , , ,,
.

=

A box contains 5 green balls and 4 red balls. Two balls will be drawn at random. How many ways to choose the same color?

add

5C z t " C z

?+.÷i
.

:÷-+¥:

to + 6 = 16 ways

Penilaian Diri Pertemuan ke 3










My Judgement to maths

Penilaian saya terhadap Pelajaran Matematika

Name : Diego Ronaldo 1.T

class : 12 Sci 4

Date : 26 October - 10th of november

NO T E S

Summary Key Points

The number ( 1) 1n . ( for a Positive number n) is called n factorial or n!
n- 2. arranging a list of n
n ! : The
. . .. .

number of ways of objects without repetition

permutation has the notation npr

describes the number of ways of choosing r objects from n
* Order does not matter

Formulas :

" Pr = n!

( ) !n - r

" Cr = n!

( ) !r !
n-r

Ka i da h Pe nc a c ah an

Aturanpenjumlahan

If a job can be completed in one way (n1) and a second job in a another way (n2) and these two tasks cannot be done simultaneously

Then the amount of ways :
hi t nz = . . . Ways

e. 9

The mathematics department will give a computer to a student or a lecturer. There are several ways to give a gift if there are 532 students and 54 lecturers

5 32 t 54 = 586 Ways

soul Aturan Penjumlahan

Mr. Adrian has 5 types of cars and 4 types of motorbikes. How many ways can Mr. Adrian travel with his personal vehicle

5 t4= 9 ways

Sharyn departs from city A to city C can have four different roads or four ways
Binar departs from city B to city C can choose 3 roads or 3 ways.
how many ways Sherin and binal meet in city C

4 t 3 = 7 ways

two six-sided dice that is (one two three four five six ), count:

the number of pairs of dice is 10

( 4,6 ) ( 5,5 ) ( 6,4 ) : Total = 3 pairs

the number of pairs of dice whose sum is at least 9

Totals : ( 6,3 )

9 ( 3. 6) ( 4,5 ) ( 514)
to (4/6) (5/5) (6/4)
11 ( 5,6 ) ( 6,5)
12 (6/6)

Total = 4 t 3 t z t 1 = 10 Pairs

In a bag, There are 10 red marbles 7 green marbles, 5 yellow marbles and 3 blue marbles. What is the probability that one marble is green, red, yellow or blue?
1- 100% = 10%

.

10

Aturan Perkalian

if there is n1 ways to do the first task and n2 ways to do the second task after the first task is done
then there is:

MI . M2 = Ways

. .. .

to finish the task

Aturan Perkalian Questions

city A and city C are connected by several roads. There are two roads through town B from city A to city B and from city B to city C there are 3 roads if a person goes from city A
to city c How many alternative roads

2. 3 = 6 ways

Teknik Filling Slot

is a method that can be used to determine the number of ways an object occupies its place

Pi Pz Pz Pn
. , .
.. .

Filling Slot Questions

there are numbers 7 5 6 4 5 will be arranged a 2-digit number

How many numbers are formed?

4 . 4 = 16 numbers

How many numbers if the numbers can't be the same

4 . 3 = 12 numbers

there are numbers 1 2 3 4 5 will be arranged to 3 digit numbers
How many numbers are formed?

5 5 5 = 125 numbers
. .

If 3 different numbers are arranged, how many numbers can be formed?
45 3 = 60 numbers

..

there are numbers 5 6 7 8 9 will be arranged 3 numbers

How many numbers are formed can be repeated

5 5 5 = 125 numbers

.

.

How many numbers if the number cannot be repeated

5 . 4 . 3 = 60 numbers

there are numbers 7 5 6 will be arranged a 2 digit number

How many odd numbers are formed?

3 2 = 6 numbers ers
.

t

7,5

How many even numbers are

3. I = 3 numbers

there are 1 2 3 4 5 numbers will be arranged 3 Numbers How many even numbers can be formed
5 . 5 . 2 = 50 numbers

there are numbers 1 2 3 4 5 will be arranged 3 Numbers how many numbers more than 300 are formed

3 5 . 5 = 75 numbers

.

345

the number of multiples of 5 consisting of three digits that can be composed of the digits 0 1 2 3 4 5 6 7 8 9 is a zero can not be in front

Otis g

9 10 . 2 = 180 numbers
.

0,5

Notasi Factorial
> Henceworth n is a natural number

n! (n facotrial) is defined as

!n = 1 2 3 . . .. . ( n - 1) n
.
.
.

n! = n 1)( !n. -

!0 = I 2! = 2I = 2 4= 1.2.3 4 = 24
1! = 3! = . .
I 6
1. 2.3 =

Permutation

Permutation With the same element 1<+1 + men
For example, from the n available elements there are k of the same elements, l of the same elements and m of the same elements (k! + l! + m! Smaller equal to than n)

then the number of permutations of the n elements is determined by

p= n!

K! l! m !

e.9

How many permutations of words “MATEMATIKA” %10.9

M = 2 units P= 10 ! 765 43T

=

A = 3 units !2 ! . 3 - 2! 1! 1! 1! 3T K
. .

..

T= 2 units n =/ 0 = 100-9 4.6.7
E= .
1 unit →

= 900 . 4.6.7

I= 1 unit = 3600.6 >
K= 1 .
unit
= 21600 7

.

= 151200

Permutation cyclic
let n different elements. The number of cyclical permutations of n elements is determined by

Pcyclical = ( n - 1) !

e. g

three people a b c sit in a circle How many ways
( n - 1) !

(3-1) ! = (2) ! = 2 ways

6 company managers around a circular table to hold a meeting How many ways can they go around the conference table in different order

( n - 1) ! = (6-1) ! = 5 ! = 5. 4. 3. 2. I = 120 Ways

Permutation with diff elements
If n and R are two positive integers and R is less than equal to r, then the number of permutations R of elements without repatriation given the notation P(n,r) is

Pln ,r) = n!

)( !n r-

The number of permutations of elements from n different elements is P(n,n) = n!

Pz> 7! 765 = 210
= =

4!

1,0¥"Pz = = 10.9 = 9-0

13 13 ! = 13.12.11 #. 1716
=
Pz = 10 ! #

7ps - bpz = 7! 6! 7.6 ! -6 !
=
- 4!
4!
4!

= ,¥6 ! (7-1) 6. != 6-5.4 6
=
= .

4! 4!

180

npz = 72 nilai n?
,
72
n! = 72

(n - 2) !

n ( n - 1) (nf) !
=

!

1)(n n - = 72

m2 n- - 72 = 0

Into )( n - g)

TM n=9

ntlp , = " Py

4+11 ! n!_

=

( ntl -3) (n- 4) !

Intl) ! n!

In )-2 ! =

In-41 !

Inti ) tnt nt
.
=

( n - 2) In - 3) (n-TH! !
(nfl ) = ( n - 2) (n - 3)

nt I = n2 - 5h 1- 6

n2 - Gn 1- 5 = 0

In - 5) ( n - 1) = 0

n=s ✓ ln -
-

t TM

74

Zntlp , - npz = 0 , nilai n?

(2nt n! 0
((2^+1)-2) !
=-

(n-3) !

{}nn? - nln - 1) in -21 In -31 ! o
=

( n - 3) !

( Intl) ( 2n ) ( 2n - 1) ! n ( n - 1) ( n -21 = 0
( 2n - 1) !
-

2)( Inti ) ( 2n)
( 1) (n- n-
n- =o

)n ( 2( zntl ) - ( n - 1) ( n - 2) = o

(2)n ( 4nF - m2 - 3h 1- 2) : o

n C - n 't Tn ) = O

( )NZn-
- Tn = 0

7)(- h2 =0
n-

n2 ( 7)n - = O

n= 7

There are 5 children who will take a photo with three or three at the coronation place for 1 2 3 winner, so many photos may be different

5 people

3 Places

5¥5ps
= = = = 60

(5-3) !

There are 4 children who will be chosen by the chairman, secretary, treasurer, how many ways to arrange?

1,4¥ 4t" P3= 24
=
=
,,

There are five children who will take a photo together on the podium 1 2 3 if one of them must always be present and occupy the 1st place, then many different photos that may occur are

Ip5- 4Pa

=

3- I

4¥ 4!¥= = 12 photos
,

Combination

combination of n elements taken from the available n elements, each element is different regardless of the order ( rsn)

ncr = n!

( )r !
!n - r

e. g

Gcy = 6 !
(6-4) ! 4 !

= 6!

2! 4 !

= 6 ✗5 4!
.

2- 4 !

= 15

"C> != it

111-71 ! > !

= 11 !

4! 7 !
10 -9?-8.7T
= 11 ..

t.3.2.TT

= 330

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