Solution for "Counting License Plates"
|
In the Kingdom of Mathlandia, all license plates consist of 3 letters, followed by 4 numbers. For example, one possible license plate is ABC-1234. Try to answer the following: 1. How many unique license plates are possible? 2. Suppose no license plate is allowed to use the letter "O". How many license plates are possible? |
|
3. Suppose no repetitions are allowed--which means that no letter or number can be used more than once. How many license plates are possible?
4. Suppose no repetitions are allowed, and the letter "M" has to be used. How many license plates are possible?
5. Suppose repetitions ARE allowed, but the sequence 211 is NOT allowed. How many license plates are possible?
To Ponder On Your Own: Would the answer to number 5 change if the disallowed sequence was 111 instead of 211?
4. Suppose no repetitions are allowed, and the letter "M" has to be used. How many license plates are possible?
5. Suppose repetitions ARE allowed, but the sequence 211 is NOT allowed. How many license plates are possible?
To Ponder On Your Own: Would the answer to number 5 change if the disallowed sequence was 111 instead of 211?
ANSWERS
1. Correct Answer: \(26^3 \times 10^4=175,760,000 \) possible license plates
How did we get this? Well, let's try to build a license plate number, starting with the first letter. Since there are 26 letters in the alphabet, there are 26 choices for the first letter, right?
Now, for each possible choice of a first letter, there are 26 choices for the second letter. So, we're already up to \(26 \times 26=26^2 \) different ways for filling in the first two letters. Continuing, we see there are \(26^3 \) different ways for filling in the 3 letters for a license plate.
But now, for each group of three letters, we have 10 choices for the first digit. And then we have 10 more choices for the second digit, 10 for the third, and 10 for the fourth. In fact, after filling in our 3 letters, we have \(10 \times 10 \times 10 \times 10=10^4\) different ways of filling in our 4 digits.
Therefore, in total, there are \(26^3 \times 10^4 \) unique license plates that are possible.
2. Correct Answer: \(25^3 \times 10^4=156,250,000\)
This restriction only affects the letter choices, and it means we only have 25 choices for each letter, rather than 26.
3. Correct Answer: \(26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7=78,624,000\)
When building our license plates this time, we still have 26 choices for the first letter. But once we've chosen that first letter, we only have 25 choices for the second letter, because we can't reuse the first letter. Then, we only have 24 choices for the third letter, because we can't reuse either of the first two letters.
Likewise for the digits, we have 10 choices for the first digit, 9 choices for the second, 8 for the third, and 7 for the fourth.
4. Correct Answer: \(3 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7=9,072,000 \)
Okay, this one's a little bit different. First, let's place the M, since we know M has to be used. There are 3 letter positions, so we have 3 choices for where to put the M.
Now, we have two more letters to choose, and they can't be M, since no repetitions are allowed. So we have \(25 \times 24\) choices for the remaining two letters.
After that, we have \(10 \times 9 \times 8 \times 7\) choices for the 4 digits. And this gives a total of \(3 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7\) unique plates.
5. Correct Answer: \(26^3\times 10^4-26^3 \times 2 \times 10=175,408,450\)
Note that repetitions are allowed, so we're back to the situation for Problem 1. But this time, we can't have the sequence 211 anywhere on the plate.
Well, the best way to do this type of problem is to find how many license plates DO have 211 somewhere, and then subtract that number from the answer to Problem 1.
So, how many ways can we build a plate with the sequence 211? First, the letters are unaffected, so there are \(26^3\) choices for the letters. Now, we have 2 choices for placing the sequence: we can use the first three digits or the last 3 digits. Once the sequence is placed, there are 10 choices for the fourth digit.
Therefore, the total number of plates WITH the sequence 211 is \(26^3\times 2 \times 10\).
This means that the total number of plates WITHOUT the sequence 211 is \(26^3 \times 10^4-26^3 \times 2 \times 10=175,408,480 \)
If you like our puzzles and explanations, please visit our store and check out our problem-solving and logic puzzle books!
