xcusemeplz2009 wrote:
There are 5 cars to be displayed in 5 parking spaces with all the cars facing A FIXED direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125
\({\text{cars}}\,\,\,\left\{ \begin{gathered}
\,3\,\,{\text{red}} \hfill \\
\,1\,\,{\text{blue}} \hfill \\
\,1\,\,{\text{yellow}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\)
\(?\,\,\,:\,\,\,\# \,\,\,{\text{in}}\,\,5\,\,{\text{parking}}\,\,{\text{spaces}}\,\,\,\,\left( {{\text{only}}\,\,{\text{colors}}\,\,{\text{matter}}} \right)\)
\(?\,\,\, = \,\,\,\underbrace {C\left( {5,3} \right)}_{{\text{red}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {2,1} \right)}_{{\text{blue}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\,\,\, = \,\,\,\,\,10 \cdot 2\,\,\, = 20\)
(Once chosen the red and blue parking spaces - among 5 and 2 available, respectively -, the yellow car is placed in the remaining parking space left.)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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