Question #257939

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


1
Expert's answer
2021-12-15T12:53:24-0500

Cn+1k=(n+1)!k!(n+1k)!C_{n + 1}^k = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}}

Cnk1+Cnk=n!(k1)!(nk+1)!+n!k!(nk)!=n!kk!(nk+1)!+n!(nk+1)k!(nk+1)!=n!k+n!(nk+1)k!(nk+1)!=n!(k+nk+1)k!(nk+1)!=n!(n+1)k!(nk+1)!=(n+1)!k!(n+1k)!C_n^{k - 1} + C_n^k = \frac{{n!}}{{(k - 1)!(n - k + 1)!}} + \frac{{n!}}{{k!(n - k)!}} = \frac{{n!k}}{{k!(n - k + 1)!}} + \frac{{n!(n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!k + n!(n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!(k + n - k + 1)}}{{k!(n - k + 1)!}} = \frac{{n!(n + 1)}}{{k!(n - k + 1)!}} = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}}

(n+1)!k!(n+1k)!=(n+1)!k!(n+1k)!Cn+1k=Cnk1+Cnk\frac{{(n + 1)!}}{{k!(n + 1 - k)!}} = \frac{{(n + 1)!}}{{k!(n + 1 - k)!}} \Rightarrow C_{n + 1}^k = C_n^{k - 1} + C_n^k


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