Question

Show that if\(n\)is a positive integer, then \(\left( {\begin{array}{*{20}{c}}{2n}\\2\end{array}} \right) = 2 \cdot \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + {n^2}\)

a) using a combinatorial argument.

b) by algebraic manipulation.

Short answer

(a) The expression\(\left( {\begin{array}{*{20}{c}}{2n}\\2\end{array}} \right) = 2 \cdot \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + {n^2}\)is proved by a combinatorial argument.

(b) The expression\(\left( {\begin{array}{*{20}{c}}{2n}\\2\end{array}} \right) = 2 \cdot \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + {n^2}\)is proved by algebraic manipulation.

Step-by-step solution

Use combinatorial argument to prove the expression

a)

Suppose a total of\(2\)objects from two boxes, each containing\(n\)of them.

This can be done in\(\left( {\begin{array}{*{20}{l}}{2n}\\2\end{array}} \right)\)no. ways.

Now choose both of them from one box or one from each box.

For the first case there are\(\left( {\begin{array}{*{20}{l}}n\\2\end{array}} \right)\)ways to do it for either of the boxes, whereas in the second case, the product rule implies that the no. of ways is \({\left( {\left( {\begin{array}{*{20}{l}}n\\1\end{array}} \right)} \right)^2}\).

Combining both:

\(\left( {\begin{array}{*{20}{c}}{2n}\\2\end{array}} \right) = 2 \cdot \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + {n^2}\)

Use an algebraic manipulation to prove the expression

b)

Use the algebraic manipulation:

\(\left( {\begin{array}{*{20}{c}}{2n}\\2\end{array}} \right) = \) Coefficient of \({x^2}\) in \({(1 + x)^{2n}} = \) Coeff. of \({x^2}\) in \({(1 + x)^n} \cdot {(1 + x)^n} = 2\).Coeff. of \({x^2}\) in \({(1 + x)^n}\). Coeff. of \({x^0}\) in \({(1 + x)^n} + \) Coeff. of \(x\) in \({(1 + x)^n}\). Coeff. of \(x\) in \({(1 + x)^n} = 2 \cdot \left( {\begin{array}{*{20}{l}}n\\2\end{array}} \right) + {\left( {\left( {\begin{array}{*{20}{l}}n\\1\end{array}} \right)} \right)^2}\).