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Combinari - exercitiu rezolvat 6

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:

=>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