Question

Kombinasi r unsur dari n unsur dinyatakan dengan nCr. Jika nC3 = 2n, nilai 2nC8 adalah....

Answer 1

NC3 = 2n
n!/[3! . (n - 3)!] = 2n
[n . (n - 1) . (n - 2) . (n - 3)!]/[3! . (n - 3)!] = 2n
[n . (n - 1) . (n - 2)]/(3!) = 2n
(n² - n) . (n - 2) = 2n . 3!
n³ - 2n² - n² + 2n = 2n . 3 . 2 . 1
n³ - 3n² + 2n = 12n
n³ - 3n² + 2n - 12n = 0
n³ - 3n² - 10n = 0
n(n² - 3n - 10) = 0
n(n² - 5n + 2n - 10) = 0
n[n(n - 5) + 2(n - 5)] = 0
n(n - 5)(n + 2) = 0
n = 0

n - 5 = 0
n = 5 (memenuhi)

n + 2 = 0
n = -2

n yang memenuhi adalah 5

2nC8
= 2(5)C8
= 10C8
= 10!/[8! . (10 - 8)!]
= (10 . 9 . 8!)/(8! . 2!)
= (10 . 9)/(2 . 1)
= 45