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Bunuel

Tough and Tricky questions: Combinations.



Since 2001, the standard serial numbers on a New York state license plate are 3 letters followed by 4 digits. How many different license plates are possible if letters and digits can be repeated?

A. 26 × 3 × 10 × 4
B. 26 × 25 × 24 × 10 × 9 × 8 × 7
C. 26³ × 9 × 9 × 9 × 9
D. 26 × 25 × 24 × 10 000
E. 26³ × 10 000

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OFFICIAL SOLUTION:

(E) The formula for permutations of events is the product of the number of ways each event can occur. There are 26 letters and 10 digits. So there are 26 × 26 × 26 options for the three letters, and 10 × 10 × 10 × 10 for the four digits. The number of different license plates is 26 × 26 × 26 × 10 × 10 × 10 × 10 = 26³ × 10 000.

The correct answer is choice (E).
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Would the answer be A if the letters and numbers were not allowed to be repeated
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Would the answer be A if the letters and numbers were not allowed to be repeated

In this case B would be correct: (26 × 25 × 24) × (10 × 9 × 8 × 7)
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Bunuel
Bunuel

Tough and Tricky questions: Combinations.



Since 2001, the standard serial numbers on a New York state license plate are 3 letters followed by 4 digits. How many different license plates are possible if letters and digits can be repeated?

A. 26 × 3 × 10 × 4
B. 26 × 25 × 24 × 10 × 9 × 8 × 7
C. 26³ × 9 × 9 × 9 × 9
D. 26 × 25 × 24 × 10 000
E. 26³ × 10 000

Kudos for a correct solution.

OFFICIAL SOLUTION:

(E) The formula for permutations of events is the product of the number of ways each event can occur. There are 26 letters and 10 digits. So there are 26 × 26 × 26 options for the three letters, and 10 × 10 × 10 × 10 for the four digits. The number of different license plates is 26 × 26 × 26 × 10 × 10 × 10 × 10 = 26³ × 10 000.

The correct answer is choice (E).

Hi,

I have a slight confusion about this problem.

Since all the answers are permutations, when we say 26 x 26 x 26 x 10 x 10 x 10 x 10 aren't we saying that order matters ?

What I mean to say is that there might very well be a number plate with CLL 1234 and as two Ls will be taken as different under this permutation based method the total count will include another CLL 1234, whereas hey should be counted as just 1 (2 number plates can not be the same).

Shouldn't we make an adjustment for this ?

Will appreciate a response on this please.
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Quote:

Hi,

I have a slight confusion about this problem.

Since all the answers are permutations, when we say 26 x 26 x 26 x 10 x 10 x 10 x 10 aren't we saying that order matters ?

What I mean to say is that there might very well be a number plate with CLL 1234 and as two Ls will be taken as different under this permutation based method the total count will include another CLL 1234, whereas hey should be counted as just 1 (2 number plates can not be the same).

Shouldn't we make an adjustment for this ?

Will appreciate a response on this please.

altairahmad
Hello, The question is asking how many combinations are possible in total. We are not concerned whether the number plates are same or not. So Lets say AAA0000 next will be AAA0001 next will be AAA0002 and so on.. So All we need to figure out is what is the number of exhaustive cases.
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