A factory makes automobile parts. Each part has a code consisting of a letter and three digits, such as C117, O076, or Z920. Last week the factory made 60,000 parts. Prove that there are at least three parts that have the same serial number.
There are 26 ways to choose the letter, and 10 ways to choose each digit for a total of 26*10³=26000 distinct serial numbers.
So there are 34000 parts that do not have distinct serial numbers. Out of the 34000 parts, we can take another 26000 to make duplicate serial numbers with the first 26000. This accounts for 52000 parts, all of which are duplicates.
To assign serial numbers to the remaining 8000 parts using the same scheme, we have to reuse at least one of the 26000 serial numbers. Therefore by the pigeon hole principle, at least three parts have the same serial number.
THANK YOU! :)
You're welcome!
Well, let's see. With 60,000 parts and each part having a code consisting of a letter and three digits, there are a total of 26 letters and 1,000 possible combinations of three digits from 000 to 999.
Now, we can use a little trickery to prove that there must be at least three parts with the same serial number. If we divide the 60,000 possible parts into groups of 1,000, that gives us 60 groups. Since there are only 26 letters in the alphabet, by the pigeonhole principle, at least two parts must have the same letter as their code.
Now, within each group of 1,000, we have 1,000 possible combinations of three digits. However, since there are 1,000 codes but only 999 unique combinations, at least two parts must have the same three-digit code within each of these groups.
Therefore, combining these two observations, we have at least two parts with the same letter and at least two parts with the same three-digit code within each of the 60 groups. This means that there must be at least two parts within each group that have the same letter-code combination.
And voila! Since there are 60 groups, and at least two parts in each group with the same letter-code combination, there must be at least 60 x 2 = 120 parts with the same serial number. So, we've proven that there are definitely at least three parts that have the same serial number!
I hope that brought a little numerical amusement to your day!
To prove that there are at least three parts with the same serial number, we can apply the Pigeonhole Principle.
Step 1: Understand the Problem
The problem states that there is a total of 60,000 parts produced in the factory, and each part has a unique serial number consisting of a letter and three digits.
Step 2: Determine the Number of Possible Serial Numbers
To find the number of possible serial numbers, we need to consider the number of options for each digit in the serial number.
For the letter, there are 26 options (A-Z).
For each digit, there are 10 options (0-9).
Therefore, the total number of possible serial numbers is 26 × 10 × 10 × 10 = 26,000.
Step 3: Divide the Total Number of Parts by the Number of Possible Serial Numbers
Dividing the total number of parts (60,000) by the number of possible serial numbers (26,000) will give us the average number of parts per serial number.
60,000 ÷ 26,000 ≈ 2.31
Since the average number is 2.31, we can conclude that on average, each serial number will be assigned to slightly more than 2 parts.
Step 4: Apply the Pigeonhole Principle
The Pigeonhole Principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain multiple pigeons.
In this case, the "pigeons" are the serial numbers (60,000 parts), and the "pigeonholes" are the possible serial numbers (26,000).
Since there are more parts (pigeons) than possible serial numbers (pigeonholes), at least one serial number (pigeonhole) must contain more than one part (pigeon).
Therefore, there must be at least three parts with the same serial number.
Step 5: Conclusion
By applying the Pigeonhole Principle, we have proven that in a factory that produced 60,000 parts with a total of 26,000 possible serial numbers, there must be at least three parts with the same serial number.