Prime divides its binomial coefficient?

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SUMMARY

The discussion centers on proving that if p is a prime number, then p divides the binomial coefficient B(p, m) for 0 < m < p. The binomial coefficient is defined as B(p, m) = p! / (m!(p-m)!). The key insight is that factoring p out of B(p, m) results in (p-1)! / (m!(p-m)!), which must be an integer. The challenge lies in demonstrating that m!(p-m)! divides (p-1)!. Participants suggest leveraging known results about divisibility to simplify the proof.

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  • Understanding of binomial coefficients and their properties
  • Familiarity with factorial notation and operations
  • Knowledge of prime numbers and their divisibility rules
  • Basic combinatorial mathematics
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  • Study the properties of binomial coefficients in combinatorics
  • Explore the concept of divisibility in number theory
  • Learn about Wilson's theorem and its implications for prime numbers
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Students in mathematics, particularly those studying combinatorics and number theory, as well as educators seeking to enhance their understanding of binomial coefficients and prime divisibility.

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Hi all, this homework problem's been driving me nuts. It seems like it's probably pretty straightforward and I'm missing something obvious, but I just can't work it out.

Homework Statement



prove that if p is a prime number that p|B(p,m) where B(p,m) is the ordinary binomial coefficient (i.e. p Choose m) for 0 < m < p

Homework Equations



B(p,m) = p!/m!(p-m)!

The Attempt at a Solution



If you factor p out of the binomial coefficient, you're left with (p-1)!/m!(p-m)!, which must be an integer. Thus I need to be able to show that m!(p-m)!|(p-1)! somehow. I've monkeyed around with the expressions to try and recover a multiple of a binomial coefficient or something, but haven't been able to... but I'm getting the feeling that I'm taking the wrong approach here =/

Any hints here would be very much appreciated!

Cheers
 
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can you show p divides (p,1)?

how about using that result to consider (p,2)?
 

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