Jawab:
1. 2! x 3! = 2 . 1 x 3 . 2 . 1 = 2 . 6 = 12
2. 6!. 4! / 5! 2 ! = 6 . 12 = 72
3. Permutasi
4. 5! = 120
5. 6! / 3! = 6 . 5 . 4 = 120
6. _ _ _
Kotak pertama hanya bisa diisi 1 dan 2
Kotak kedua dan ketiga bisa diisi semua bilangan
maka 2 . 5 . 5 = 50 bilangan
7. 6! / 3! = 6 . 5 . 4 = 120 pemilihan yang mungkin
1. 2! × 3!
= ( 2 × 1 ) × ( 3 × 2 × 1 )
= 2 × 6
= 12
2. 6! . 4! / 5! . 2!
= (6.5.4.3.2.1) . (4.3.2.1) / (5.4.3.2.1) . (2.1)
= 720 . 24 / 120 . 2
= 17.280/240
= 72
3. kaidah pencacahan / permutasian
4. SENAR
S = 1
E = 1
N = 1
A = 1
R = 1
===== +
TOTAL = 4!
GANDA = 0
= 4!
= (4.3.2.1)
= 24 SUSUNAN
5. AMANDA
A = 3
M = 1
D = 1
TOTAL = 6!
GANDA = 3!
= 6!/3!
= (6.5.4.3.2.1)/(3.2.1)
= 720/6
= 120 SUSUNAN
6. 2 × 5 × 5
= ( 2 × 5 ) × 5
= 10 × 5
= 50
7. 6! ÷ 3!
= (6.5.4.3.2.1) ÷ (3.2.1)
= 720 ÷ 3
= 120 pemilihan yg mungkin terjadi
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Jawab:
1. 2! x 3! = 2 . 1 x 3 . 2 . 1 = 2 . 6 = 12
2. 6!. 4! / 5! 2 ! = 6 . 12 = 72
3. Permutasi
4. 5! = 120
5. 6! / 3! = 6 . 5 . 4 = 120
6. _ _ _
Kotak pertama hanya bisa diisi 1 dan 2
Kotak kedua dan ketiga bisa diisi semua bilangan
maka 2 . 5 . 5 = 50 bilangan
7. 6! / 3! = 6 . 5 . 4 = 120 pemilihan yang mungkin
1. 2! × 3!
= ( 2 × 1 ) × ( 3 × 2 × 1 )
= 2 × 6
= 12
2. 6! . 4! / 5! . 2!
= (6.5.4.3.2.1) . (4.3.2.1) / (5.4.3.2.1) . (2.1)
= 720 . 24 / 120 . 2
= 17.280/240
= 72
3. kaidah pencacahan / permutasian
4. SENAR
S = 1
E = 1
N = 1
A = 1
R = 1
===== +
TOTAL = 4!
GANDA = 0
= 4!
= (4.3.2.1)
= 24 SUSUNAN
5. AMANDA
A = 3
M = 1
N = 1
D = 1
===== +
TOTAL = 6!
GANDA = 3!
= 6!/3!
= (6.5.4.3.2.1)/(3.2.1)
= 720/6
= 120 SUSUNAN
6. 2 × 5 × 5
= ( 2 × 5 ) × 5
= 10 × 5
= 50
7. 6! ÷ 3!
= (6.5.4.3.2.1) ÷ (3.2.1)
= 720 ÷ 3
= 120 pemilihan yg mungkin terjadi