AMC8 1999

AMC8 1999 · Q15

AMC8 1999 · Q15. It mainly tests Basic counting (rules of product/sum).

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set ${C,H,L,P,R}$, the second from ${A,I,O}$, and the third from ${D,M,N,T}$. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?

Flatville的自行车车牌每个包含三个字母。第一位从集合${C,H,L,P,R}$选，第二位从${A,I,O}$选，第三位从${D,M,N,T}$选。为了需要更多车牌，他们添加了两个新字母。新字母可都加到一个集合，或各加到一个集合。添加两个字母能制造的最大可能新增车牌数是多少？

(A) 24 24

(B) 30 30

(D) 40 40

(E) 60 60

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that $5 \cdot 3 \cdot 4 = 60$ plates could have been made. If two letters are added to the second set, then $5 \cdot 5 \cdot 4 = 100$ plates can be made. If one letter is added to each of the second and third sets, then $5 \cdot 4 \cdot 5 = 100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100 - 60 = 40$ ADDITIONAL plates can be made. Note: Optimum results can usually be obtained in such problems by making the factors as nearly equal as possible.

答案（D）：在添加新字母之前，第一个位置可以从5个不同字母中选择，第二个位置从3个中选择，第三个位置从4个中选择。这意味着可以制作 $5 \cdot 3 \cdot 4 = 60$ 个牌照。如果在第二组中增加两个字母，那么可以制作 $5 \cdot 5 \cdot 4 = 100$ 个牌照。如果在第二组和第三组各增加一个字母，那么可以制作 $5 \cdot 4 \cdot 5 = 100$ 个牌照。其余四种放置这两个字母的方法都不会产生这么多牌照。因此，$100 - 60 = 40$ 个额外的牌照可以制作。注：在这类问题中，通常通过让各因数尽可能接近来获得最优结果。

Topics

CP.001 Basic counting (rules of product/sum)