Jeonghee has 5 different red cards and 5 different blue card

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

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

The correct answer is (C).

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

Regards,
Fabio.

=>

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.

Therefore, the answer is C.
Answer: C