# Prove: C(n, k) = C(n, n-k). using theorem C(n, k) = n! / {k!(n-k)!}?

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- 1 decade agoBest Answer
C(n,k) = n! / {(k!)(n-k)!}

C(n,n-k) = n! / {(n-k)!(n-(n-k))!} = n! / {(n-k)!(k!)}

C(n,n-k) = C(n,k)

- 1 decade ago
C(n,k) = C(n, n-k)

C(n, n-k) = n!/ {(n-k)![n-(n-k)]!}

= n!/{(n-k)![n-n+k]!}

= n!/{(n-k)!k!} = C(n, k)

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