3. Hitunglah :
a. 1/(6 ! )+ 2/(8 !)- 3/(9 !) = . . .
b. C(n,2 ) = 105 ,tentukan nilai n
c. 10 P(n,4) = P(n,5) ,tentukan nilai n
Jawaban :
a. 1/(6 ! )+ 2/(8 !)- 3/(9 !)
= 1/6! + 2/(8 . 7 . 6!) - 3/(9 . 8 . 7 . 6!)
= (9 . 8 . 7)/(9 . 8 . 7 . 6!) + (2 . 9)/(9 . 8 . 7 . 6!) - 3/(9 . 8 . 7 . 6!)
= ((9 . 8 . 7)+ (2 . 9) - 3) /(9 . 8 . 7 . 6!)
= 519 / (9 . 8 . 7 . 6!)
= 519/9!
= 519/362880
b. C(n,2 ) = 105 ,tentukan nilai n
nC2 = 105
n!/2!(n - 2)! = 105
n . (n - 1) . (n - 2)! / (2 . (n - 2)!) = 105
n . (n - 1) / 2 = 105
n . (n - 1) = 210
n^2 - n = 210
n^2 - n - 210 = 0
(n - 15)(n + 14) = 0
n = 15 atau n = 14
karena n > 0, maka n yang dipilih : n = 15
c. 10 P(n,4) = P(n,5) ,tentukan nilai n
10 . nP4 = nP5
10 . n! / (n - 4)! = n! / (n - 5!)
10 / (n - 4)! = 1/ (n - 5)!
10 . (n - 5)! = (n - 4) !
10 . (n - 5)! = (n - 4) . (n - 5) !
10 = n - 4
10 + 4 = n
n = 14
No comments:
Post a Comment