Масъала.
Айниятро исбот кунед:
$$C_n^k C_{n-k}^{m-k}=C_m^k C_n^m\quad\quad (*).$$

Ҳал.
\(C_n^m = \frac{n!}{(n - m)! m!}\),
ки дар ин ҷо \(m \leq n; C_n^0 = 1;\)
\(C_n^k = \frac{n!}{(n - k)! k!}\)
\(C_{n-k}^{m-k} = \frac{(n-k)!}{(n - k-(m-k))! (m-k)!}=\frac{(n-k)!}{(n - k-m+k)!(m-k)!}=\frac{(n-k)!}{(n-m)!(m-k)!}\)
\(C_m^k = \frac{m!}{(m - k)! k!}\)
\(C_n^m = \frac{n!}{(n - m)! m!}\)

\begin{multline}
C_n^k C_{n-k}^{m-k}=\frac{n!}{(n - k)! k!}\cdot \frac{(n-k)!}{(n-m)!(m-k)!}=\\
=\frac{n!}{(n-m)!(m-k)!k!}=\frac{n!}{k!(n-m)!(m-k)!}\quad (1).
\end{multline}

\begin{multline}
C_m^k C_n^m=\frac{m!}{(m - k)! k!}\cdot \frac{n!}{(n - m)! m!}=\\
=\frac{n!}{(m-k)!k!(n-m)!}=\frac{n!}{k!(n-m)!(m-k)!}\quad (2).
\end{multline}

Аз баробариҳои (1) ва (2) мебарояд, ки баробарии (*) дуруст аст.

Айният исбот шуд.