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Sehingga, dengan demikian:
2 .
3.
4.
..
..
n.
1. nC0 --> 1
2. nC1 --> n , n=2 --> 2
3. nC2 --> 1/2 n (n-1) , n= 3 --> 1/2.3(3-1) = 3
4. nC3 --> 1/6 n (n-1)(n-2), n = 4 --? 1/6.4 (4-1)(4-2) = 4
.
.
n. nCn --> 1
sehigga berlaku segitiga pascal
1 1
1 1 2
1 2 1 4
1 3 3 1 8
1 4 6 4 1 16
dst
1,2,4,8,16,... sehingga mengikutipola