求助NN几道prep maths 急！

acc

    

39.   

    

A certain stock
exchange designates each stock with a one-, two-, or three-letter code, where
each letter is selected from the

    

26 letters of the
alphabet.  If the letters may be repeated
and if the same letters used in a different order constitute a different code,
how many different stocks is it possible to uniquely designate with these
codes?

    

 

    

(A)  2,951

    

(B)  8,125

    

(C)  15,600

    

(D)  16,302

    

(E)  18,278

    

 

    

E

    

41.  

    

All of the stocks on
the over-the-counter market are designated by either a 4-letter or a 5-letter
code that is created by using the 26 letters of the alphabet.  Which of the following gives the maximum
number of different stocks that can be designated with these codes?

    

 

    

(A) 2(26^5)

    

(B) 26(26^4)

    

(C) 27(26^4)

    

(D) 26(26^5)

    

(E) 27(26^5)

    

 

    

C

    

43.   

    

A committee of
three people is to be chosen from four married couples.  What is the number of different committees
that can be chosen if two people who are married to each other cannot both
serve on the committee?

    

 

    

(A)  16

    

(B)  24

    

(C)  26

    

(D)  30

    

(E)  32

    

 

    

E

多年不碰数学了，这几题的答案怎么也算不出，向各位XDJM求教，非常感谢！！

AlexLam

1:

组合问题：1位数的时候为26个选项中选一，即C(26)1；2位数的时候，因为数字可以重复，所以个位和十位都是C(26)1，如此类推。计算式：

C(26)1+C(26)1*C(26)1+C(26)1*C(26)1*C(26)1=26+26^2+26^3＝18，278

2：

同上，计算式：

26^4+26^5=26^4(1+26)=27(26^4)

3：

没有couple的组合＝所有组合－有couple的组合，计算式：

所有组合＝C(8)3＝56

有couple的组合＝C(4)1*C(6)1=24: C(4)1是4对couple选一对，C(6)1是剩下6个人里选一个。

答案＝56－24＝32。

acc

感动，没想到这么快就有回复 ，谢谢！谢谢！
AlexLam写得很清楚，看得很明白