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3. Permutasi
Permutasi adalah banyaknya susunan (cara pengurutan) dari semua atau sebagian unsur.
a. Permutasi n unsur yang berbeda
Contoh:
1) Tentukan banyaknya cara untuk menyusun huruf-huruf H, A, T, I.
Penyelesaian
n = 4
4 P 4 = 4! = 4 . 3. 2. 1 = 24 cara
2) Tentukan banyaknya bilangan yang dapat dibentuk dari angka 5, 7, dan 9 jika
tidak boleh ada angka yang sama.
Penyelesaian
n = 3
3 P 3 = 3! = 3. 2. 1 = 6 cara
b. Permutasi r unsur dari n unsur yang berbeda
Keterangan
n : banyaknya unsur yang berbeda
r : unsur yang diambil
Contoh:
1) Hitunglah nilai dari 5 P 3 !
Penyelesaian
5 P 3 =
5!
(5−3)!
=
5!
2!
=
5.4.3.2!
2!
= 5. 4. 3 = 60
2) Tentukan banyak susunan 3 huruf dari huruf-huruf A, B, C, D, E, dan F.
Penyelesaian
n = 6
r = 3
6 P 3 =
6!
(6−3)!
=
6!
3!
=
6.5.4.3!
3!
= 6. 5. 4 = 120 susunan
n P n = n . (n – 1) . (n – 2) . ... 3. 2. 1 = n!
n P r =
𝑛!
(𝑛−𝑟)!
, 𝑟 ≤ 𝑛
c. Permutasi yang memuat unsur yang sama
Contoh:
Berapa kata dapat disusun dengan semua huruf pada kata “MATEMATIKA” !
Penyelesaian
n = 10
k1 = M = 2, k2 = A = 3, k3 = T = 2
10 P 2,3,2 =
10!
2!3!2!
=
10.9.8.7.6.5.4.3!
2!3!2!
=
10.9.8.7.6.5.4
4
= 151.200
d. Permutasi Siklis
Permutasi siklis adalah permutasi yang memperhatikan tempat kedudukan unsur di
lingkaran (ditempatkan secara melingkar).
Contoh:
1. Hitung banyak permutasi siklis yang terdiri atas 7 unsur yang berlainan.
Penyelesaian
7 P (siklis) = (7 − 1)! = 6! = 6.5.4.3.2.1 = 720
2. Dalam suatu pertemuan, ada 8 peserta akan menempati 8 kursi yang mengelilingi
meja bundar. Berapa banyak susunan yang mungkin terjadi?
Penyelesaian
8 P (siklis) = (8 − 1)! = 7! = 7.6.5.4.3.2.1 = 5.040 susunan
e. Permutasi Berulang
Contoh:
1. Berapa banyak susunan 3 huruf yang diambil dari huruf-huruf “K, A, M, I, dan
S” , jika unsur-unsur yang tersedia itu boleh ditulis berulang?
Penyelesaian
n = 5, r = 3
P (berulang) = 53
= 125 susunan
2. Berapa banyak bilangan terdiri 2 angka yang dapat disusun dari angka-angka 3,
4, 6, 7, 8, dan 9, jika angka-angka yang tersedia boleh ditulis berulang?
Penyelesaian
n = 6, r = 2
P (berulang) = 62
= 36 susunan
n P (k1, k2, ...., kn) =
𝑛!
𝑘1! 𝑘2! 𝑘3 !… 𝑘𝑛!
n P (siklis) = (𝑛 − 1)!
P (berulang) = 𝑛𝑟

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Peluang pert 2

  • 1. 3. Permutasi Permutasi adalah banyaknya susunan (cara pengurutan) dari semua atau sebagian unsur. a. Permutasi n unsur yang berbeda Contoh: 1) Tentukan banyaknya cara untuk menyusun huruf-huruf H, A, T, I. Penyelesaian n = 4 4 P 4 = 4! = 4 . 3. 2. 1 = 24 cara 2) Tentukan banyaknya bilangan yang dapat dibentuk dari angka 5, 7, dan 9 jika tidak boleh ada angka yang sama. Penyelesaian n = 3 3 P 3 = 3! = 3. 2. 1 = 6 cara b. Permutasi r unsur dari n unsur yang berbeda Keterangan n : banyaknya unsur yang berbeda r : unsur yang diambil Contoh: 1) Hitunglah nilai dari 5 P 3 ! Penyelesaian 5 P 3 = 5! (5−3)! = 5! 2! = 5.4.3.2! 2! = 5. 4. 3 = 60 2) Tentukan banyak susunan 3 huruf dari huruf-huruf A, B, C, D, E, dan F. Penyelesaian n = 6 r = 3 6 P 3 = 6! (6−3)! = 6! 3! = 6.5.4.3! 3! = 6. 5. 4 = 120 susunan n P n = n . (n – 1) . (n – 2) . ... 3. 2. 1 = n! n P r = 𝑛! (𝑛−𝑟)! , 𝑟 ≤ 𝑛
  • 2. c. Permutasi yang memuat unsur yang sama Contoh: Berapa kata dapat disusun dengan semua huruf pada kata “MATEMATIKA” ! Penyelesaian n = 10 k1 = M = 2, k2 = A = 3, k3 = T = 2 10 P 2,3,2 = 10! 2!3!2! = 10.9.8.7.6.5.4.3! 2!3!2! = 10.9.8.7.6.5.4 4 = 151.200 d. Permutasi Siklis Permutasi siklis adalah permutasi yang memperhatikan tempat kedudukan unsur di lingkaran (ditempatkan secara melingkar). Contoh: 1. Hitung banyak permutasi siklis yang terdiri atas 7 unsur yang berlainan. Penyelesaian 7 P (siklis) = (7 − 1)! = 6! = 6.5.4.3.2.1 = 720 2. Dalam suatu pertemuan, ada 8 peserta akan menempati 8 kursi yang mengelilingi meja bundar. Berapa banyak susunan yang mungkin terjadi? Penyelesaian 8 P (siklis) = (8 − 1)! = 7! = 7.6.5.4.3.2.1 = 5.040 susunan e. Permutasi Berulang Contoh: 1. Berapa banyak susunan 3 huruf yang diambil dari huruf-huruf “K, A, M, I, dan S” , jika unsur-unsur yang tersedia itu boleh ditulis berulang? Penyelesaian n = 5, r = 3 P (berulang) = 53 = 125 susunan 2. Berapa banyak bilangan terdiri 2 angka yang dapat disusun dari angka-angka 3, 4, 6, 7, 8, dan 9, jika angka-angka yang tersedia boleh ditulis berulang? Penyelesaian n = 6, r = 2 P (berulang) = 62 = 36 susunan n P (k1, k2, ...., kn) = 𝑛! 𝑘1! 𝑘2! 𝑘3 !… 𝑘𝑛! n P (siklis) = (𝑛 − 1)! P (berulang) = 𝑛𝑟