# Prove that two non-bald residents of NYC have exactly the same number of hairs.

In New York City there are two non-bald people who have the same number of hairs ( the human head can contain up to several hundred thousands with maximum of about 500,000)

How can I prove the pigeonhole principle of the two non-bald people in New York City ?

The pigeonhole principle: if k is a positive integer and k+1 or more objects are placed into k boxes then there is at least one box containing two or more of the objects.

• How many people are there in New York? What if they all had different numbers of hairs? – MJD Jul 3 '15 at 4:56
• @Betty, you already had an existing account which you were using a few minutes ago. Why did you create a new account? – Joel Reyes Noche Jul 3 '15 at 5:01
• Also, you forgot to include in the question the information that you already know. Explain to the users what your idea of the pigeonhole principle is. Then find out how many people there are in New York (Try typing "population of New York City" in Wolfram Alpha.) then put that in your question. – Joel Reyes Noche Jul 3 '15 at 5:11
• Sorry I didn't remember my password, I try to delete and ask the question again. – Betty Jul 3 '15 at 5:11
• Don't delete this question any more. There is already an answer and it would be rude to the answerer if you delete the question. Just go back to your original account for your next question. – Joel Reyes Noche Jul 3 '15 at 5:13

• You actually have to assume that there are more than $500,000$ people with hair on their head. But that's not very difficult, since you would only need that more than $5\%$ of the people of New York has hair in their heads, which seems like a safe assumption. – Arthur Jul 3 '15 at 5:16