Question #255838

Show that C (n+1, k) = C (n, k -1) + C (n, k)


1
Expert's answer
2021-10-25T15:29:26-0400
C(n+1,k)=(n+1k)=(n+1)!k!(n+1k)!C(n+1,k)=\dbinom{n+1}{k}=\dfrac{(n+1)!}{k!(n+1-k)!}

C(n,k1)=(nk1)=n!(k1)!(nk+1)!C(n,k-1)=\dbinom{n}{k-1}=\dfrac{n!}{(k-1)!(n-k+1)!}

C(n,k)=(nk)=n!k!(nk)!C(n,k)=\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}

Then


C(n,k1)+C(n,k)C(n,k-1)+C(n,k)

=n!(k1)!(nk+1)!+n!k!(nk)!=\dfrac{n!}{(k-1)!(n-k+1)!}+\dfrac{n!}{k!(n-k)!}

=n!(k+nk+1)k!(nk+1)!=n!(n+1)k!(nk+1)!=\dfrac{n!(k+n-k+1)}{k!(n-k+1)!}=\dfrac{n!(n+1)}{k!(n-k+1)!}

=(n+1)!k!(n+1k)!=C(n+1,k)=\dfrac{(n+1)!}{k!(n+1-k)!}=C(n+1,k)




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