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

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

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New post 20 Mar 2019, 01:58
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Question Stats:

47% (02:07) correct 53% (02:53) wrong based on 15 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?

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

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Re: Jeonghee has 5 different red cards and 5 different blue cards. She shu  [#permalink]

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New post 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.
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Jeonghee has 5 different red cards and 5 different blue cards. She shu  [#permalink]

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New post 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
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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.
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Re: Jeonghee has 5 different red cards and 5 different blue cards. She shu  [#permalink]

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New post 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
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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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