permutasi
No. 1
P(n,5) = 10 P(n,4)
n!/(n-5)! = 10 . n!/(n-4)!
10 = (n-4)! / (n-5)!
10 = (n-4)(n-5)! / (n-5)!
10 = n - 4
n = 14
No. 2
P(n,2) = 72
n!/(n-2)! = 72
n(n-1)(n-2)!/(n-2)! = 72
n(n-1) = 9 × 8
n = 9
No. 3
P(n+1,3) = P(n,4)
(n+1)! /(n+1-3)! = n!/(n-4)!
(n+1)!/(n-2)! = n!(n-4)!
(n+1)n(n-1) = n(n-1)(n-2)(n-3)
n+1 = n² - 5n + 6
n² - 6n + 5 = 0
(n-1)(n-5) = 0
n = 5
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Verified answer
permutasi
No. 1
P(n,5) = 10 P(n,4)
n!/(n-5)! = 10 . n!/(n-4)!
10 = (n-4)! / (n-5)!
10 = (n-4)(n-5)! / (n-5)!
10 = n - 4
n = 14
No. 2
P(n,2) = 72
n!/(n-2)! = 72
n(n-1)(n-2)!/(n-2)! = 72
n(n-1) = 9 × 8
n = 9
No. 3
P(n+1,3) = P(n,4)
(n+1)! /(n+1-3)! = n!/(n-4)!
(n+1)!/(n-2)! = n!(n-4)!
(n+1)n(n-1) = n(n-1)(n-2)(n-3)
n+1 = n² - 5n + 6
n² - 6n + 5 = 0
(n-1)(n-5) = 0
n = 5