Category Archives: Soal dan Jawaban Matematika Diskrit

Correct me if I’m wrong. Thank you :)

Finite Automata


Note dan latihan :

Finite Automata

Sirkuit euler, hamiltonian, trees


Note :

sirkuit euler,hamiltonian,isomorphic,trees

Latihan :

Latihan euler, graph dkk

Relasi, Refleksi, Simetri, Asimetri, Transitif, Poset, Hasse Diagram


catatan relasi,refleksi,simetri,asimetri,transitf,poset

Latihan :

Latihan seputar relasi dkk

Induksi Matematika


Note :

This is the first time I have to search the solution in google because my lecturer didn’t teach this problem (2n +1 < 2^n, n>=3).

Catatan tentang Fuzzy Set


Saya buatkan dalam bentuk pdf

Catatan fuzzy set :

Fuzzy set

Latihan Fuzzy Set  :

Latihan Fuzzy Set

Latihan kedua fuzzy set :

Latihan Fuzzy Set pt2

Catatan dan Latihan mencari rekursif


20121218_185929

 

 

 

 

 

 

 

 

 

 

 

Tentukan fungsi rekursif dari bentuk  berikut:

  1. Fungsi pangkat n,  ,n integer
  2. Deret : 1+2+3+4+…+n
  3. Deret : 2+4+6+8+10+…+n
  4. Barisan bilangan    , n integer

20121218_190026

Latihan matdis sebelum UTS



20121118_184206

Contoh soal binomial theorem


Tentukan koefisien x13  pada binomial [ ( x3 +1 )2  ( x2 -  )8 ]

[ ( x3 +1 )2  ( x2 -  )8 ] = [ ( x3 +1 )2  ( x2 – 2x-1 )8 ]

(x3 +1)2  = x6  + 2x3 + 1

( x2 – 2x-1 )8  = [ x2 + ( – 2x-1 ) ]8

Misalkan : a = x2 , b = (– 2x-1 )

(a+b) 8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8

= (x2)8 + 8 (x2)7 (– 2x-1 ) + 28 (x2)6 (– 2x-1 )2 + 56 (x2)5 (– 2x-1 )3 + 70 (x2)4

(– 2x-1 )4 + 56 (x2)3 (– 2x-1 )5 + 28 (x2)2 (– 2x-1 )6 + 8(x2) (– 2x-1 )7 +

(– 2x-1 )8

=  x16 +(-16) x13 + (28*4) x10 + [ 56 *(-8) x7 ] + …………….

[ ( x3 +1 )2  ( x2 – 2x-1 )8 ]

x6                x7         =  1 * 56 * (-8) = – 448

x3                   x10        =  2 * 28 * 4     =   224

x0                x13        = 1 * (-16)        =   – 16  +

koefisien x13 = – 240 .

inklusi-eksklusi pt 3


Soal :
•Dengan prinsip inklusi eksklusi.tentukan banyaknya bilangan prima yang tidak melebihi 100
Jawab :

  akar 100 = 10, bilangan prima yang kurang dari 10 = 2,3,5,7 (n=4)

Banyaknya bilangan yang tidak lebih dari 100 = 4 + banyak bilangan bulat postif antara 100 yang habis dibagi 2,3,5, atau 7 = 4 + (A U B U C U D)

A= habis dibagi 2 = 100 / 2 = 50.

B =habis dibagi 3 = 100 / 3 = 30.

C =habis dibagi 5 = 100 / 5 = 20.

D = habis dibagi 7 = 100 / 7 = 14.

A n B = habis dibagi 2 dan 3 = 100 / 6 = 16.

A n C = habis dibagi 2 dan 5 = 100 / 10 = 10.

A n D = habis dibagi 2 dan 7 = 100 / 14 = 7.

B n C = habis dibagi 3 dan 5 = 100 / 15 = 6.

B n D = habis dibagi 3 dan 7 = 100 / 21 = 4.

C n D = habis dibagi 5 dan 7 = 100 / 35 = 2.

A n B n C = habis dibagi 2,3, dan 5 = 100 / 30 = 3.

A n B n D = habis dibagi 2,3, dan 7 = 100 / 42 = 2.

A n C n D = habis dibagi 2,5, dan 7 = 100 / 70 = 1.

B n C n D = habis dibagi 3,5, dan 7 = 100 / 105 = 0.

A n B n C n D = habis dibagi 2,3,5, dan 7 = 100 /105 = 0.

Banyak bilangan bulat yang tidak melebihi 100 = 99.

Rumus :

|A U B U C U D| = |A|+|B|+|C|+|D| – |A n B| – |A n C| +

|A n D|- |B n C | – |B n C|- |C n D| + |A n B n C| + |A n B n D| +

|A n C n D| + |B n C n D| – |A n B n C n D|

|A U B U C U D| = 99 – (50 + 33 + 20 + 14 – 16 – 10 – 7 – 6 – 4 – 2 + 3 + 2 + 1 + 0 + 0 )

= 99 – 78 =  21.

Banyaknya bilangan prima yang tidak melebihi 100 = 4 + 21 = 25.

inklusi-eksklusi pt 2


Soal :
Notasi | A| menyatakan banyaknya elemen himpunan  A. Terdapat sebuah himpunan semesta S dan tiga buah himpunan bagiannya yaitu himpunan X, himpunan Y dan himpunan Z, dimana :

Bila ,

a. Tentukan nilai p

b. Tentukan banyaknya elemen masing-masing himpunan S, X, Y, dan Z.

Jawab :

|X| = p + 5 + 2p = 3p + 5

|Y| = 2 ( 3p + 5 ) +1 = 6p + 11

|Z| = 10 + 1 + p = 11 + p

|X U Y U Z | =
9p – 20 = (p+5) + 2p + (3p + 10) + (1 + p) + 10
9p – 20 = 7p + 26
2p = 46
p = 23

|X| = p + 5 + 2p = 3p + 5 = 3 * 23 +5 = 74

|Y| = 2 ( 3p + 5 ) +1 = 6p + 11 = 6 * 23 + 11 = 149

|Z| = 10 + 1 + p = 11 + p = 11 + 23 = 34

|S| = 9p = 9 * 23 = 207.

 

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