Maths problem: 我好質疑（c)的計法, （d)也不同意,可否賜教, thanks?

(c) 前兩類任選2: C(6+8,2) = 14×13÷2 = 91
     第一類取1, 後兩類取1: 6×(4+6) = 60
     第二類取1, 第三類取1: 8×4 = 32
     合計 91+60+32 = 183.


(d) 取1: 6+8+4+6 = 24
     取2: (c) 之答案: 183
     取3: 只能從第一類取: C(6,3) = 20
     合計: 24+183+20 =227


從一堆 n 不同物取 k 件, 方法數是 C(n,k)
   = [n(n-1)...(n-k+1)]/[k(k-1)....1]
不是 n(n-1)...(n-k+1), 因為會重複算.
例如從 A,B,C 3物取2, 只有 3×2÷2 = 3 種方法
(不計順序的話). 因為: 譬如取的是 A,B, 考慮順
序的話有先取A再取B, 或先取B再取A. 3×2 算
的是第一次取有 3種選擇, 第二次取還有 2種選
擇, 先A後B 和 先B後A 是兩種取法. 但如果不計
順序, 只問: 取了什麼, 則上述兩種都是取了 A,B,
只能算一種. 所以3物取2物的取法只有 AB, AC,
BC 3種.

上面的 solution :-
=============
(c) Methods of buying 2 books using ≦ $50 are:
{16+16,16+20,16+28,16+34, 20+20, 20+28} 


∴ no. of ways buying 2 books using ≦ $50 
= P(6,2) + 6*8+6*4+6*6+ P(8,2) +8*4 - - - - - (X) 
　^^^^^　　　　　　　　^^^^^^
= 30　　+48+24+36+　56　　+32 - - - - - - (X) 
= 226 - - - - - - (X) 


REASON:
6 different books(A,B,C,D,E,F), choose 2 : 
(A,B),(A,C),(A,D),(A,E),(A,F),
(B,A),(B,C),(B,D),(B,E),(B,F),
(C,A),(C,B),(C,D)(C,E),(C,F),
(D,A), . . . . . . . . . . . . .,(D,F),
(E,A), . . . . . . . . . . . . .,(E,F),
(F,A), . . . . . . . . . . . . . ,(F,E)
∴ no. of ways =  P(6,2) = 6*5 = 30

But, when choosing 2 books, (A,B)=(B,A), (A,C)=(C,A), . . .
☆☆ no. of ways SHOULD = C(6,2) = (6*5)/2 = 15  

The answer for (c) SHOULD BE : 
C(6,2) + 6*8+6*4+6*6+ C(8,2) +8*4 = 183 
________________________________________________
(d) Methods of buying 1 book using ≦ $50 are:
{16, 20, 28, 34} 
∴ no. of ways buying 1 book using ≦ $50 
= 6+8+4+6 - - - - - - - - - - - - (✓) 
= 24 

Method of buying 3 books using ≦ $50 is:
{16+16+16} 
∴ no. of ways buying 3 books using ≦ $50 
= P(6,3) - - - - - - - - - - - - (X), SHOULD BE C(6,3)
= 120 - - - - - - - - - - - - (X), SHOULD BE 20

∴ no. of ways of buying any no. of books using ≦ $50 
= 24+226+120 = 370  - - - - - - - - - - - - (X) 

☆☆The answer SHOULD BE : 24+183+20 = 227