數學:P和C

Question

1.為何132*(2n-2)!/n!(n-2)!=35*(2n)!/(n-1)!(n+1)!會變成
132(n-1)(n+1)=35(2n)(2n-1)?
2.為何n!/(k-1)!(n-K+1)!:n!/k!(n-k)!:n!/(k+1)!(n-k+1)!=k(k+1):(k+1)(n-k+1):(n-k+1)(n-k)
3.k>2,k屬於N,C43取k+1=C43取2k,求Pk取2=?(請寫出計算過程)
4.Pn取m=60,Cn取m=10,m,n,屬於N,求m`n=?(請寫出計算過程)

Answer 1

1.為何132*(2n-2)!/n!(n-2)!=35*(2n)!/(n-1)!(n+1)!會變成
132(n-1)(n+1)=35(2n)(2n-1)?
132*(2n-2)! 35* (2n)! 132*(2n-2)! 35*(2n)(2n-1)(2n-2)!
--------------- = ----------------- => ----------------- = ---------------------------
n! * (n-2)! (n-1)! * (n+1)! n(n-1)!*(n-2)! (n-1)!*(n+1)n(n-1)(n-2)!
=> ( 交叉相乘) 132(n-1)(n+1) = 35(2n)(2n-1)
=> (乘開) 132n2 - 132 = 140n2 -70n
=> (移項) 8n2 - 70n + 132 = 0
=> (約分) 4n2 - 35n + 66 = 0
2. 題目有錯 (n-k+1)! 應該是 (n-k-1)!
為何n!/(k-1)!(n-K+1)!:n!/k!(n-k)!:n!/(k+1)!(n-k-1)!=k(k+1):(k+1)(n-k+1):(n-k+1)(n-k)
n! n! n!
----------------- : --------- : ----------------
(k-1)!(n-k+1)! k!(n-k)! (k+1)!(n-k-1)!
(n-k+1)! =(n-k+1)(n-k)(n-k-1)!
k! = k(k-1)!; (n-k)! = (n-k)(n-k-1)!
(k+1)! = (k+1)(k)(k-1)!
分子同時消去 n!, 分母同時消去 (k-1)!(n-k-1)!
1 1 1
=> --------------- : -------- : ---------
(n-k+1)(n-k) k(n-k) (k+1)k
(同時乘以 (n-k+1)(n-k)(k+1)k )
=> k(k+1) : (k+1)(n-k+1) : (n-k+1)(n-k)
3. k>2, k屬於N, C(43, k+1) = C(43, 2k)
因為 C(m, n) = C(m, m-n)
=> C(43, 2k) = C(43, 43-2k) = C(43, k+1)
=> k+1 = 2k 或 43-2k = k+1
=> k = 1 或 k = 14
=> k=1 時 P(1, 2) = 0 (無解)
k = 14 時 P(14, 2) = 91
4. P(n, m) = 60, C(n, m) = 10, m,n,屬於N
60 P(n, m) n!/(n-m)!
--- = --------- = ------------------- = m!
10 C(n, m) n!/[(n-m)!*m!]
=> m! = 6
=> m = 3
=> P(n, 3) = [n(n-1)(n-2)(n-3)!] / (n-3)! = 60
=> n(n-1)(n-2) = 60
因為這是排列組合 n 直接用數字代比較快
又 n > ｍ 所以從 4 開始
4 => 4*3*2 = 24 ( x )
5 => 5*4*3 = 60 ( O )
m = 3, n = 5
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.