標題:
數學題(Combination)#9 10分
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00145646/o/701112150033413873411940.jpg ans. 38: r=10,n=29 11a=3 b=9 c=4 請列出步驟,thanks
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最佳解答:
38 The r + 1 th term is C(n,r), The r th term is C(n,r - 1) The r + 10 th term is C(n,r + 9), The r + 11 th term is C(n,r + 10) n!/[r!(n - r)!] = 2n!/[(r - 1)!(n - r + 1)!]...(1) n!/[(r + 9)!(n - r - 9)!] = 2n!/[(r + 10)!(n - r - 10)!]...(2) So, 2r = n - r + 1...(3) 2(n - r - 9) = r + 10...(4) Solve it, 4r - 20 = r + 10 3r = 30 r = 10 => n = 29 11(a) A->D->E->G ; A->D->F->G ; A->B->C->F->G 3 routes (b) 3 * 3 = 9 routes (c) E->D ; F->D ; G->E->D ; G->F->D 4 return routes
其他解答:
38. general (x+y)^n = summation of nCw [x^(n-w)] [y^w] (1+x)^n = summation of nCw [1^(n-w)] [x^w] (r+1)th term => w = r the coefficient at (r+1)th term = nCr rth term => w = r-1 the coefficient at rth term = nCr-1 (r+10)th term => w = r+9 the coefficient at rth term = nCr+9 (r+11)th term => w = r+10 the coefficient at rth term = nCr+10 nCr = 2 nCr-1 n! / [r!(n-r)!] = 2(n!) / [(r-1)!(n-r+1)!] 1/ r = 2/ (n-r+1) 2r = n-r+1 => 3r = n + 1 ... (1) nCr+9 = 2 nCr+10 n! / [(r+9)!(n-r-9)!] = 2(n!) / [(r+10)!(n-r-10)!] 1/ (n-r-9) = 2/ (r+10) r + 10 = 2n-2r-18 => 3r = 2n - 28 ...(2) sub (1) into (2) n + 1 = 2n - 28 n = 29 => r = 10 11a. (without backtracking supposes no city to be reached twice) A>B>C>F>D>E>G A>B>C>F>G A>B>D>E>G A>B>D>F>G so it should be 4 routes from city A to city G 11b. from city A to city G has 4 possible routes from city G to city A has 4 possible routes so total is 4x4 = 16 possible routes from city A to city G and back to city A 11c. consider return routes only there are 3 possible routes on return travel through city D
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