Question

Use Exercise 33 to give an alternative proof of Corollary 2 in Section 6.3, which states that \(\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}n\\{n - k}\end{array}} \right)\) whenever k is an integer with 0 ≤ k ≤ n. (Hint: Consider the number of paths of the type described in Exercise 33 from (0, 0) to (n − k, k) and from (0, 0) to (k, n − k).)

Short answer

\(\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}n\\{n - k}\end{array}} \right)\)

Step-by-step solution

Step 1: Considerations

Consider the two journeys viz.

\(\begin{array}{l}{P_1} = \;from\;(0,0)\,to\;(k,n - k)\\{P_2} = \;from\;(0,0)\,to\;(n - k,k)\end{array}\)

Step 2: Symmetry

From symmetry, the number of different paths for both

\(\begin{array}{l}{P_1} = \;from\;(0,0)\,to\;(k,n - k)\\{P_2} = \;from\;(0,0)\,to\;(n - k,k)\end{array}\)

Are the same.

This implies using the result of 33

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{k + (n - k)}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{(n - k) + k}\\k\end{array}} \right)\\\left( \begin{array}{l}n\\k\end{array} \right) = \left( {\begin{array}{*{20}{c}}n\\{n - k}\end{array}} \right)\end{array}\)