Bantu nomer 3 yang b yaa

nC3 + nC4 = (n+1)C3

n!/3!(n-3)! + n!/4!(n-4)! = (n+1)!/3!(n+1-3)!

n(n-1)(n-2)(n-3)!/3×2×1(n-3)! + n(n-1)(n-2)(n-3)(n-4)!/4×3×2×1(n-4)! = (n+1)(n+1-1)(n+1-2)(n+1-3)!/3×2×1(n+1-3)!

{n(n-1)(n-2)/6} + {n(n-1)(n-2)(n-3)/24} = {(n+1)(n)(n-1)/6}
kalikan 24

4n(n-1)(n-2) + n(n-1)(n-2)(n-3) = 4n(n+1)(n-1)
karena n dan n - 1 ada di semua berarti dicoret maka sisa

4(n-2) + (n-2)(n-3) = 4(n+1)
4n - 8 + n² - 5n + 6 = 4n + 4
n² - n - 2 - 4n - 4 = 0
n² - 5n - 6 = 0
(n - 6)(n + 1) = 0
n = 6 atau n = -1