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Prove that: $$p \,\,\left|\, {p \choose k} \right., \quad 0< k \lt p$$ if $p$ is prime.

how to prove that with direct proof?

Henry T. Horton
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World
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  • Hint: think of the explicit definition of $\displaystyle\binom{p}{k}$ as a fraction. How many times does $p$ divide the numerator? How many can it divide the denominator? – Steven Stadnicki Nov 28 '12 at 18:40
  • can you explain more? – World Nov 28 '12 at 18:45
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    Be careful that $k>0$ since $\binom{p}{0}=1$ and then it is not true that $p|1$. – Sebastien B Nov 28 '12 at 18:48
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    Also k < p for the same reason. – Gautam Shenoy Nov 28 '12 at 18:51
  • @GautamShenoy You are right, I had not seen that both the inequalities were large in the title. – Sebastien B Nov 28 '12 at 19:41
  • Combinatorially, the set ${X\choose k}$ of $k$-subsets of an $n$-element set $X$ ($1<k<n$) can be partitioned into $n$ disjoint equal-sized cells. The cells are $\Gamma_x=\left{V:x\in V\in{X\choose k}\right}$ for $x\in X$, which we may intuitively expect to be a partition by considering symmetry. – anon Nov 28 '12 at 20:45

1 Answers1

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Write out what the binomial coefficient is: $$ {p\choose k}=\frac{p!}{k!(p-k)!}. $$ $p$ divides the numerator since it has a factor of $p$, but $p$ can't divide the denominator because it is the product of integers smaller than $p$ and $p$ is prime.

This means that $p$ does not appear in the prime factorisation of the denominator, thus you can't simplify the $p$ factor that is on the numerator.

Jean-Sébastien
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