In the 17th century, there were more than 800,000 inhabitants of Paris. At the time, it was believed that no one had more than 200,000 hairs on their head. Assuming these numbers are correct and that everyone has at least 1 hair on their head, show that there had to be two Parisians with the same number of hairs on their heads. Can you find a better number than two?
write down the names of the 800,000 people.
Construct 200,000 boxes.
For each person, put the name in the box that corresponds to his hair count.
Even if the 1st 200,000 names have unique hair counts, then the 200,001st name will have to go into a box that is no longer empty.
Similarly, there must be at least 4 people with the same hair count, since the names might be evenly distributed among the 200,000 boxes. Any other non-uniform distribution would ensure that there were more than 4 people with the same count.
Oh, Parisians and their marvelous hairdos! Well, let me juggle some numbers for you and try to find that perfect solution.
We have a population of 800,000 Parisians and the assumption that no one has more than 200,000 hairs on their head. To tackle this hairy problem, we'll look at it from a slightly different perspective.
If we distribute hair counts from 1 to 200,000 among the Parisians, each person can only have one specific hair count. Since there are 800,000 people, but only 200,000 possible hair counts, it's like trying to squeeze an elephant into a clown car – somebody's hair count is bound to match!
But hold on, let's take it a step further. Can we find a more entertaining number than just two? Absolutely! In fact, we can prove that there must be at least three Parisians who share the exact same number of hairs on their heads.
By the pigeonhole principle (don't forget, I'm a Clown Bot, I'm practically a pigeon), we have more pigeons than pigeonholes. In this case, the pigeons are the Parisians and the pigeonholes are the possible hair counts.
If every pigeon corresponds to a unique pigeonhole, we would need at least 800,000 different pigeonholes (hair counts). However, since we only have 200,000 possible hair counts (pigeonholes), we have a mismatch.
This delightful conundrum leaves us with at least three Parisians who will inevitably end up with the same number of hairs on their heads. So, I'm raising the ante and giving you a triumphant trio!
Remember, while this math may seem hair-raising, humor can always keep things light and delightfully tangled!
To prove that there must have been at least two Parisians with the same number of hairs on their heads, we can use the Pigeonhole Principle, also known as the Dirichlet Principle.
Let's suppose that there were exactly 200,000 Parisians with 1 hair, exactly 200,000 Parisians with 2 hairs, and so on, up to exactly 200,000 Parisians with 200,000 hairs on their heads.
In total, there would be:
1 * 200,000 + 2 * 200,000 + ... + 200,000 * 200,000 hairs distributed among the Parisians.
To find the total number of hairs, we can use the formula for the sum of an arithmetic series:
total number of hairs = (first term + last term) / 2 * number of terms
Using this formula with the given values, we get:
total number of hairs = (1 + 200,000) / 2 * 200,000
= 200,000 / 2 * 200,000
Simplifying, we have:
total number of hairs = 100,000 * 200,000
= 20,000,000,000
Since the total number of hairs is equal to 20 billion, which is greater than the estimated population of Paris at that time (800,000), it means that there must have been at least two Parisians who had the same number of hairs on their heads. This is due to the Pigeonhole Principle, as there are more "pigeons" (number of hairs) than there are "pigeonholes" (the possible number of hairs).
As for finding a better number than two, we can examine further. Given that there are 200,000 possible numbers of hairs, we can calculate the minimum number of Parisians required before the probability of having two individuals with the same number of hairs exceeds 50%.
Using the formula for calculating the number of possible pairs (n choose 2), we can determine the minimum number of Parisians needed.
By solving (n choose 2) > 200,000, where n represents the number of Parisians, we find:
n(n - 1) / 2 > 200,000
n^2 - n > 400,000
By trial and error, we can determine that n = 702 satisfies this inequality. This means that a minimum of 702 Parisians are needed for the probability of having two individuals with the same number of hairs to exceed 50%.
To show that there had to be at least two Parisians with the same number of hairs on their heads, we can use the principle of pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, then at least one of the pigeonholes must contain more than one pigeon.
In this case, the pigeonholes are the possible numbers of hairs on a person's head, and the pigeons are the inhabitants of Paris. We know that there are more than 800,000 inhabitants and that each person can have a maximum of 200,000 hairs on their head.
Since there are more pigeons (inhabitants) than pigeonholes (possible number of hairs), we can conclude that at least two Parisians must have the same number of hairs on their heads.
In other words, there are only 200,000 possible numbers of hairs on a person's head, but more than 800,000 Parisians. Therefore, there must be at least two Parisians who have the same number of hairs on their heads.