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20 Mar 2019, 01:58
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C
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Difficulty:
95%
(hard)
Question Stats:
51% (02:52) correct
49%
(02:50)
wrong
based on 134
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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?
\(A. \frac{2}{5}\)
\(B. \frac{28}{125}\)
\(C. \frac{31}{126}\)
\(D. \frac{33}{140}\)
\(E. \frac{25}{216}\)
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?
\(A. \frac{2}{5}\)
\(B. \frac{28}{125}\)
\(C. \frac{31}{126}\)
\(D. \frac{33}{140}\)
\(E. \frac{25}{216}\)
20 Mar 2019, 08:35
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MathRevolution
[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)!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?
\(A. \frac{2}{5}\)
\(B. \frac{28}{125}\)
\(C. \frac{31}{126}\)
\(D. \frac{33}{140}\)
\(E. \frac{25}{216}\)
\(? = P\left( {{\rm{red}}\,\,{\rm{together}}\,{\rm{,}}\,{\rm{blue}}\,\,{\rm{together}}} \right)\)
\({\rm{total}}\,:\,\,\,\underbrace {C\left( {10,5} \right)}_{{\rm{cards}}\,\,{\rm{chosen}}}\,\, \cdot \,\,\underbrace {\,\,5!\,\,}_{{\rm{cards}}\,\,{\rm{chosen}}\,\,{\rm{order}}}\,\,\,\, = \,\,\,{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} \over {5 \cdot 4 \cdot 3 \cdot 2}}\,\,\, = \,\,\,2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!\,\,\,\,{\rm{equiprobable}}\,\,{\rm{sequences}}\)
\(\left. \matrix{\\
\left( 1 \right)\,\,\left\{ \matrix{\\
\,5\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,5\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,{\rm{5!}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 5! = 240 \hfill \cr \\
\left( 2 \right)\,\,\left\{ \matrix{\\
\,4\,\,{\rm{red}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,4\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{1}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,4} \right) \cdot {\rm{4!}}\,\, \cdot {\rm{C}}\left( {5,1} \right)} \right] = 2 \cdot 25 \cdot 24 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 24 = 2400 \hfill \cr \\
\left( 3 \right)\,\,\left\{ \matrix{\\
\,3\,\,{\rm{red}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{blue}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr \\
\,{\rm{or}} \hfill \cr \\
\,3\,\,{\rm{blue}}\,{\rm{,}}\,{\rm{2}}\,\,{\rm{red}}\,\,{\rm{chosen}}\,\,{\rm{:}}\,\,\,2!\, \cdot \,\left[ {C\left( {5,3} \right) \cdot {\rm{3!}}\,\, \cdot {\rm{C}}\left( {5,2} \right) \cdot 2!} \right] = 2 \cdot 10 \cdot 6 \cdot 10 \cdot 2 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,100 \cdot 48 = 4800 \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,7440\)
\(? = {{7440} \over {2 \cdot 3 \cdot 7 \cdot 6 \cdot 5!}} = \ldots = {{31} \over {126}}\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Updated on: 22 Mar 2019, 01:21
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.
Last edited by Archit3110 on 22 Mar 2019, 01:21, edited 1 time in total.
Kudos
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MathRevolution
[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?
\(A. \frac{2}{5}\)
\(B. \frac{28}{125}\)
\(C. \frac{31}{126}\)
\(D. \frac{33}{140}\)
\(E. \frac{25}{216}\)
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
