1999 AMC 8 Problems/Problem 15

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Problem

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?

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution

Solution 1

There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.

Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.

Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.

Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.

Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.

Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.

Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.

You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed {D}$

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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