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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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Jeonghee has 5 different red cards and 5 different blue cards. She shu  [#permalink]

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20 Mar 2019, 01:58
1
3
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Difficulty:

55% (hard)

Question Stats:

57% (02:25) correct 43% (02:46) wrong based on 21 sessions

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[GMAT math practice question]

Jeonghee has 5 different red cards and 5 different blue cards. She shuffles the 10 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?

$$A. \frac{2}{5}$$
$$B. \frac{28}{125}$$
$$C. \frac{31}{126}$$
$$D. \frac{33}{140}$$
$$E. \frac{25}{216}$$

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: Jeonghee has 5 different red cards and 5 different blue cards. She shu [#permalink] ### Show Tags 20 Mar 2019, 08:35 MathRevolution wrote: [GMAT math practice question] Jeonghee has 5 different red cards and 5 different blue cards. She shuffles the 10 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? $$A. \frac{2}{5}$$ $$B. \frac{28}{125}$$ $$C. \frac{31}{126}$$ $$D. \frac{33}{140}$$ $$E. \frac{25}{216}$$ 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. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net GMAT Club Legend Joined: 18 Aug 2017 Posts: 5436 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Jeonghee has 5 different red cards and 5 different blue cards. She shu [#permalink] ### Show Tags Updated on: 22 Mar 2019, 01:21 MathRevolution wrote: [GMAT math practice question] Jeonghee has 5 different red cards and 5 different blue cards. She shuffles the 10 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? $$A. \frac{2}{5}$$ $$B. \frac{28}{125}$$ $$C. \frac{31}{126}$$ $$D. \frac{33}{140}$$ $$E. \frac{25}{216}$$ so as to get 5 cards out of 10 , P would be 10*9*8*7*6 case 1: BBBBB; RRRRR ; 5*4*3*2*1 case 2: BBBBR, RBBBB, RRRRB and BRRRR ; 5∗4∗3∗2∗55∗4∗3∗2∗5 case 3 ; BBBRR, RRBBB, RRRBB and BBRRR ; 5∗4∗3∗5∗4 total Probability : (5∗4∗3∗2∗1)∗2+(5∗4∗3∗2∗5)∗4+(5∗4∗3∗5∗4)∗4/ 10*9*8*7*6 solve we get 31/126 IMO C Originally posted by Archit3110 on 20 Mar 2019, 23:58. Last edited by Archit3110 on 22 Mar 2019, 01:21, edited 1 time in total. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8235 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Jeonghee has 5 different red cards and 5 different blue cards. She shu [#permalink] ### Show Tags 22 Mar 2019, 00:37 => 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 $$\frac{( 5*4*3*2*1*2 + 5*4*3*2*5*4 + 5*4*3*5*4*4 )}{10*9*8*7*6} = \frac{{ 5*4*3(4+40+80) }}{{ 10*9*8*7*6 }} = \frac{124}{2*9*2*7*2} = \frac{31}{126}$$. Therefore, the answer is C. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. 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Re: Jeonghee has 5 different red cards and 5 different blue cards. She shu   [#permalink] 22 Mar 2019, 00:37
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