C 3n + C 4n = n(n - 2)
combinari de n luate cate 3 + combinari de n luate cate 4 = n(n-2) n = ?
Rezolvare
se aplica formula combinarilor: combinari de n luate cate k si rezulta:
se aplica formula combinarilor: combinari de n luate cate k si rezulta:
=>n! / 3!(n - 3)! + n! / 4!(n - 4)! = n (n - 2)
=> (n - 3)! (n - 2) (n - 1) n / 3!(n - 3)! + (n - 4)! (n -
3) (n - 2) (n - 1) n / 4!(n - 4)!
= n(n - 2)
=> (*4) (n - 2)
(n - 1) n / 3! + (n - 3) (n - 2) (n -
1) n / 4! = n(n - 2)
=> 4n (n - 1)(n - 2)
+ (n - 3) (n - 2) (n - 1) n = 4!
* n(n - 2)
=> n(n - 1)(n - 2)[3n + n - 3] = 1 * 2 * 3 * 4 * n(n –
2) / (n – 2)
=> n(n - 1)[4n - 3] = 24n
=> n2 – n + 4n2 – 3n = 24n
=> 5n2 – 4n = 24n
=> 5n2 – 4n - 24n = 0
=> 5n2 – 28n = 0
=> n(5n - 28) = 0
a) n = 0
5n – 28 = 0