permutasian dari kata
-bey
-off
-dulu
-mau
-tidur
note; saudara kalalawar
Quizz
bey
total unsur : 3
unsur ganda : –
3!
3×2×1
= 6 susunan
off
total unsur : 3
unsur ganda : 2
3! / 2!
3×2×1 / 2×1
= 6 / 2
= 3 susunan
dulu
total unsur : 4
unsur ganda : 2
4! / 2!
4×3×2×1 / 2×1
= 24 / 2
= 12 susunan
mau
total unsur : 3
unsur ganda : –
3!
3×2×1
= 6 susunan
tidur
total unsur : 5
unsur ganda : –
5!
5×4×3×2×1
= 120 susunan
Jawaban:
Penyelesaian
• Bey
B = 1
E = 1
Y = 1
Jumlah huruf = 3
p = n!
= 3!
= 3.2.1
= 6 Susunan
• Off
O = 1
F = 2
Jumlah huruf = 3
Unsur ganda = 2
p = n! / k!
= 3! / 2!
= 3.2.1 / 2.1
= 6 / 2
= 3 susunan
• Dulu
D = 1
U = 2
L = 1
Jumlah huruf = 4
Unsur ganda = 2
p = n! / k!
= 4.3.2.1 / 2.1
= 24 / 2
= 12 susunan
• Mau
M = 1
A = 1
U = 1
Jumlah huruf = 3
p = n!
= 3!
= 3.2.1
= 6 susunan
• Tidur
T = 1
I = 1
D = 1
U = 1
R = 1
Jumlah huruf = 5
p = n!
= 5!
= 5.4.3.2.1
= 120 susunan
Detail Jawaban
- Mapel : Matematika
- Kelas : 12
- Materi : Bab 7 – Kaidah Pencacahan
- Kode soal : 2
- Kode kategorisasi : 12.2.7