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Prove that: C ( n , k ) = P ( n , k ) r ! \displaystyle{C(n, k)=\frac{P(n, k)}{r!}}

Notation:

* C ( n , k ) C(n, k) is a Combination

* P ( n , k ) P(n, k) is a Permutation

n ! n ! ( n − k ) ! = n ! k ! ( n − k ) ! \frac{n!}{n!(n-k)!}=\frac{n!}{k!(n-k)!} n ! n ! ( k − n ) ! = k ! n ! ( n − k ) ! \frac{n!}{n!(k-n)!}=\frac{k!}{n!(n-k)!} n ! k ! ( n − k ) ! = n ! k ! ( n − k ) ! \frac{n!}{k!(n-k)!}=\frac{n!}{k!(n-k)!} n ! k ! ( n − k ) ! = n ! n ! ( n − k ) ! \frac{n!}{k!(n-k)!}=\frac{n!}{n!(n-k)!} n ! k ! ( n − k ) ! = n ! k ! ( n − k ) ! \frac{n!}{k!(n-k)!}=\frac{n!}{k!(n-k)!}

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1 solution

The value of permutation is P ( n , k ) = n ! ( n − k ) ! P(n, k) = \frac{n!}{(n-k)!} and also the combination is C ( n , k ) = n ! k ! ( n − k ) ! C(n, k) = \frac{n!}{k!(n-k)!}

ADIOS!!! \large \text{ADIOS!!!}

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