nC3+nC2=165點計?
given nCr+nC(r-1) = (n+1)Cr
maths
2010-06-01 4:01 am
回答 (2)
2010-06-01 4:08 am
✔ 最佳答案
nC3 + nC2 = 165By given nCr+nC(r-1) = (n+1)Cr ,
(n+1)C3 = 165
(n+1)! / [3! (n+1 - 3)!] = 165
(n+1)! / (n - 2)! = 165 * 3!
(n - 1) n (n + 1) = 990
(n - 1) n (n + 1) = 9 * 10 * 11
n = 10
2010-06-04 9:08 am
given that nCr+nC(r-1) = (n+1)Cr
nC3+nC2 = 165
nC3+nC(3-1) = 165
(n+1)C3 = 165
(n+1)! / [3!(n+1-3)!] = 165
(n+1)! / (n-2)! = 165 x 3!
[(n+1)(n)(n-1)(n-2)(n-3)...(3)(2)(1)] / [(n-2)(n-3)...(3)(2)(1)] = 990
(n+1)(n)(n-1) = 9 x 10 x 11
n = 9
nC3+nC2 = 165
nC3+nC(3-1) = 165
(n+1)C3 = 165
(n+1)! / [3!(n+1-3)!] = 165
(n+1)! / (n-2)! = 165 x 3!
[(n+1)(n)(n-1)(n-2)(n-3)...(3)(2)(1)] / [(n-2)(n-3)...(3)(2)(1)] = 990
(n+1)(n)(n-1) = 9 x 10 x 11
n = 9
參考: me
收錄日期: 2021-04-21 22:16:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100531000051KK01336