ATIKA 5+3 poin Jika nCr menyatakan banyaknya kombinasi r elemen dari n elemen dan nC3 = 2n maka 2nC7

NC3 = 2n
=> n! / (n-3)!.3! = 2n
=> n (n – 1)(n – 2)(n – 3)! / (n – 3)!.3.2.1 = 2n
=> n (n^2 – 3n + 2) / 6 = 2n
=> n^2 – 3n + 2 = 12
=> n^2 – 3n – 10 = 0
=> (n – 5)(n + 2) = 0
=> n = 5 atau n = -2
yang diambil n = 5
2nC7 = 10C7 = 10!/(10-7)!7! = 10.9.8.7!/3.2.1.7!= 10.3.4= 120