f.5數學 – 排列與組合

2012-03-19 4:08 am

回答 (1)

2012-03-19 4:42 am
✔ 最佳答案
9(a) P(n,2) = n(n - 1)

P(n + 2,4) = (n + 2)(n + 1)n(n - 1)

P(n + 1,3) = (n + 1)^2 n^2 (n - 1)^2

P(n,2)P(n + 2,4) - P(n + 1,3)P(n + 1,3)

= n(n - 1)(n + 2)(n + 1)n(n - 1) - (n + 1)^2 n^2 (n - 1)^2

= (n - 1)^2 n^2(n + 1)(n + 2 - n - 1)

= (n - 1)^2 n^2 (n + 1)

(b) C(n,r - 1) + C(n,r)

= n!/(r - 1)!(n - r + 1)! + n!/r!(n - r)!

= [n(n - r + 1) + nr]/(r)!(n - r + 1)!

= n(n + 1)/(r)!(n - r + 1)!

[C(n,r - 1) + C(n,r) ]/P(n,r - 1) * r!

= [n(n + 1)/(r)!(n - r + 1)!]/P(n,r - 1) * r!

= (n + 1)/(n - 1)!





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