Jeonghee has 5 different red cards and 5 different blue cards. She shu

Beautiful problem, Max. Congrats (and kudos)!


\(? = P\left( {{\rm{red}}\,\,{\rm{together}}\,{\rm{,}}\,{\rm{blue}}\,\,{\rm{together}}} \right)\)


\({\rm{total}}\,:\,\,\,\underbrace {C\left( {10,5} \right)}_{{\rm{cards}}\,\,{\rm{chosen}}}\,\, \cdot \,\,\underbrace {\,\,5!\,\,}_{{\rm{cards}}\,\,{\rm{chosen}}\,\,{\rm{order}}}\,\,\,\, = \,\,\,{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} \over {5 \cdot 4 \cdot 3 \cdot 2}}\,\,\, = \,\,\,2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!\,\,\,\,{\rm{equiprobable}}\,\,{\rm{sequences}}\)


\(\left. \matrix{\\
\left( 1 \right)\,\,\left\{ \matrix{\\
\,5\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,5\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 5! = 240 \hfill \cr \\
\left( 2 \right)\,\,\left\{ \matrix{\\
\,4\,\,{\rm{red}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,4\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 24 = 2400 \hfill \cr \\
\left( 3 \right)\,\,\left\{ \matrix{\\
\,3\,\,{\rm{red}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,3\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 48 = 4800 \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,7440\)


\(? = {{7440} \over {2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!}} = \ldots = {{31} \over {126}}\)


The correct answer is (C).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.