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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Jeonghee has 5 different red cards and 5 different blue card tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## Jeonghee has 5 different red cards and 5 different blue card ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [GMAT math practice question] Jeonghee has 5 different red cards and 5 different blue cards. She shuffles the10 cards, and then places 5 of the cards in a row. What is the probability that all red cards are adjacent to each other and all blue cards are adjacent to each other in her row? A. 2/5 B. 28/125 C. 31/126 D. 33/140 E. 25/216 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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Max@Math Revolution wrote:
[GMAT math practice question]

Jeonghee has 5 different red cards and 5 different blue cards. She shuffles the10 cards, and then places 5 of the cards in a row. What is the probability that all red cards are adjacent to each other and all blue cards are adjacent to each other in her row?

A. 2/5
B. 28/125
C. 31/126
D. 33/140
E. 25/216
$$? = P\left( {{\rm{red}}\,\,{\rm{together}}\,{\rm{,}}\,{\rm{blue}}\,\,{\rm{together}}} \right)$$
$${\rm{total}}\,:\,\,\,\underbrace {C\left( {10,5} \right)}_{{\rm{cards}}\,\,{\rm{chosen}}}\,\, \cdot \,\,\underbrace {\,\,5!\,\,}_{{\rm{cards}}\,\,{\rm{chosen}}\,\,{\rm{order}}}\,\,\,\, = \,\,\,{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} \over {5 \cdot 4 \cdot 3 \cdot 2}}\,\,\, = \,\,\,2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!\,\,\,\,{\rm{equiprobable}}\,\,{\rm{sequences}}$$
$$\left. \matrix{ \left( 1 \right)\,\,\left\{ \matrix{ \,5\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr \,{\rm{or}} \hfill \cr \,5\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 5! = 240 \hfill \cr \left( 2 \right)\,\,\left\{ \matrix{ \,4\,\,{\rm{red}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr \,{\rm{or}} \hfill \cr \,4\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 24 = 2400 \hfill \cr \left( 3 \right)\,\,\left\{ \matrix{ \,3\,\,{\rm{red}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr \,{\rm{or}} \hfill \cr \,3\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 48 = 4800 \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,7440$$
$$? = {{7440} \over {2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!}} = \ldots = {{31} \over {126}}$$

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

Regards,
Fabio.

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### GMAT/MBA Expert

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=>

The total number of ways in which 5 cards can be chosen out of 10 cards is 10P5 = 10*9*8*7*6.

There are 5*4*3*2*1 arrangements of each of BBBBB and RRRRR.
There are 5*4*3*2*5 arrangements of each of BBBBR, RBBBB, RRRRB and BRRRR.
There are 5*4*3*5*4 arrangements of each of BBBRR, RRBBB, RRRBB and BBRRR.

Thus, the total number of arrangements with all red cards adjacent to each other and all blue cards adjacent to each other is (5*4*3*2*1)*2 + (5*4*3*2*5)*4 + (5*4*3*5*4)*4.
The required probability is ( 5*4*3*2*1*2 + 5*4*3*2*5*4 + 5*4*3*5*4*4 ) / 10*9*8*7*6 = { 5*4*3(4+40+80) } / { 10*9*8*7*6 } = 124 / 2*9*2*7*2 = 31/126.

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