Solve-it-in-N-C-n-3-C-n-2-5-n-3-6n-2-6-

Question Number 87732 by mathocean1 last updated on 05/Apr/20

Solve it in N

C_{n}^{3} - C_{n}^{2} = 5 + \frac{n^{3} - {6 n}^{2}}{6}

Commented by Tony Lin last updated on 06/Apr/20

\frac{n (n - 1) (n - 2)}{6} - \frac{n (n - 1)}{2} = \frac{30 + n^{3} - 6 n^{2}}{6}

n (n - 1) (n - 2) - 3 n (n - 1) = 30 + n^{3} - 6 n^{2}

(n^{3} - 3 n^{2} + 2 n) - 3 n^{2} + 3 n = 30 + n^{3} - 6 n^{2}

5 n = 30 \Rightarrow n = 6