GWD 17 - 2

sunshine001

I can't find any posts on this question. Hope this is not a re-post.

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetic order.  What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A.    4

B.    5

C.    6

D.    7

E.     8

Ans = B

My answer is A. Maybe I didn't get through the question. My understanding is: can use SINGLE letter or 2 letters. So only need 4 alphabets. 4 letters + 4*3 > 12

Please help!

laihong

x + C(2,x)>12 ->x at least is 5

carolhide

以下是引用laihong在2005-11-21 18:05:00的发言：
x + C(2,x)>12 ->x at least is 5

能解释下么？

hitlzc

我觉得是这样吧：关键是对in alphabetic order的理解。

原来我是这样想的，可以这样排,如AB、BA、BC

如果选B的话，肯定不能BA这样排。

如果只用两个字母的话，就是C(X,2)>12,X>=4

既用一个字母又用两个字母，就如一楼所算

X大于等于5即可。

[此贴子已经被作者于2005-12-18 12:45:23编辑过]

aph7

原来如此，忽略了in alphabetic order，哎又被文字游戏蒙骗。。。

Zita

我的思路是取A开始算起,因为是LEAST嘛,很快知道是B了,各位见笑了...

yellowmonkey

Why it is not P2,5. since it has to be in sequence?

wildmantomba

我的算法是这样的
假设有X个字母
则X个字母的组合方式如下

只选1个字母：X种方式
选两个字母：X（X-1）/2种方式

有12个人
所以，最小的X应该为
X+X（X-1）/2 >=12
X（X+1）>=12
X最小值为5（因为5×6=30>24 而4×5=20<24）