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There are 5 cars to be displayed in 5 parking spaces with all the cars
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16 Dec 2009, 22:53
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There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible? (A) 20 (B) 25 (C) 40 (D) 60 (E) 125
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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10 Oct 2018, 11:02
xcusemeplz2009 wrote: There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125 Let R, R, R, B, Y represent the cars (by their colors) Notice that the three R's are identical. So, the question becomes In how many different ways can we arrange the letters R, R, R, B and Y?ASIDE When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this: If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....] So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows: There are 11 letters in total There are 4 identical I's There are 4 identical S's There are 2 identical P's So, the total number of possible arrangements = 11!/[( 4!)( 4!)( 2!)] ONTO THE QUESTION! We have R, R, R, B and Y: There are 5 letters in total There are 3 identical R's So, the total number of possible arrangements = 5!/( 3!) = (5)(4)(3)(2)(1)/(3)(2)(1) = 20 Answer: A Cheers, Brent
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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16 Dec 2009, 23:45
I'm going A.
I think you just think of it as 2 different cars and 5 empty spots, since the rest have to be red anyway. Then there's 5 spaces for the yellow car and four spaces for the blue one for each yellow car space. 4x5 = 20
Probably if I could remember the maths from permutations and combinations you'd end up with something like 5!/3!



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There are 5 cars to be displayed in 5 parking spaces with all the cars
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17 Dec 2009, 02:19
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?(A) 20 (B) 25 (C) 40 (D) 60 (E) 125 I think the above logic is correct except one thing: # of arrangements is \(\frac{5!}{3!}=20\). Answer: A. Though I think that the question is quite ambiguous. We are told that "all the cars are facing the SAME direction", not that they all are facing ONE specific direction. That's why I think we should multiply by 2.
A car can be arranged facing TWO directions (within one particular parking space), so we should multiply 20 by 2, \(20*2=40\)
They can be arranged as > > > > > 20 combinations.
AND < < < < < 20 combinations.
Stem just says that they can can not be arranged for example like this: < > > < <
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There are 5 cars to be displayed in 5 parking spaces with all the cars
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10 Oct 2018, 14:22
xcusemeplz2009 wrote: There are 5 cars to be displayed in 5 parking spaces with all the cars facing A FIXED direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125
\({\text{cars}}\,\,\,\left\{ \begin{gathered} \,3\,\,{\text{red}} \hfill \\ \,1\,\,{\text{blue}} \hfill \\ \,1\,\,{\text{yellow}} \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\) \(?\,\,\,:\,\,\,\# \,\,\,{\text{in}}\,\,5\,\,{\text{parking}}\,\,{\text{spaces}}\,\,\,\,\left( {{\text{only}}\,\,{\text{colors}}\,\,{\text{matter}}} \right)\) \(?\,\,\, = \,\,\,\underbrace {C\left( {5,3} \right)}_{{\text{red}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {2,1} \right)}_{{\text{blue}}\,\,{\text{parking}}\,\,{\text{choices}}}\,\,\,\,\, = \,\,\,\,\,10 \cdot 2\,\,\, = 20\) (Once chosen the red and blue parking spaces  among 5 and 2 available, respectively , the yellow car is placed in the remaining parking space left.) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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13 Oct 2018, 18:28
xcusemeplz2009 wrote: There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125 The cars can be displayed in the following number of ways, using the indistinguishable permutations formula: (5!)/(3!) = 5 x 4 = 20 ways. Note that we divided by 3! because the 3 red cars are identical to each other (i.e., they are indistinguishable). Answer: A
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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14 Oct 2018, 02:32
I went about this a slightly longer way;
3 Red cars, 5 spaces > 15 possibilities 1 Blue car, 5 spaces > 5 possibilities 1 Yellow car, 5 spaces > 5 possibilities
Total number of arrangements: 15+5+5 = 20.



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There are 5 cars to be displayed in 5 parking spaces with all the cars
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19 Jan 2019, 03:35
swatato wrote: I went about this a slightly longer way;
3 Red cars, 5 spaces > 15 possibilities 1 Blue car, 5 spaces > 5 possibilities 1 Yellow car, 5 spaces > 5 possibilities
Total number of arrangements: 15+5+5 = 20. no .... 15+5+5 <> 20



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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12 Mar 2019, 12:04
Bunuel wrote: There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?(A) 20 (B) 25 (C) 40 (D) 60 (E) 125 I think the above logic is correct except one thing: # of arrangements is \(\frac{5!}{3!}=20\). Answer: A. Though I think that the question is quite ambiguous. We are told that "all the cars are facing the SAME direction", not that they all are facing ONE specific direction. That's why I think we should multiply by 2.
A car can be arranged facing TWO directions (within one particular parking space), so we should multiply 20 by 2, \(20*2=40\)
They can be arranged as > > > > > 20 combinations.
AND < < < < < 20 combinations.
Stem just says that they can can not be arranged for example like this: < > > < < What about the "front / back" disposition, as far as I can tell we have the 20 possibilities with all cars pointed to the frond and 20 extra with them pointed to the back?



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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12 Mar 2019, 13:23
Possible no. of arrangements = 5!/3! = 5*4 = 20 i.e A



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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18 May 2019, 04:24
Hi
Can anyone please explain, why the answer here is not 40. (5!/3!)*2 In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements.



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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18 May 2019, 05:13
vinayakvaish wrote: Hi
Can anyone please explain, why the answer here is not 40. (5!/3!)*2 In a parking slot all cars can face 2 directions right? and these 2 directions will comprise of 2 different arrangements. Question says they all face the same direction.



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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18 May 2019, 07:43
all cars can face either front or back. question says one direction but does not specify a direction



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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18 May 2019, 09:06
vinayakvaish wrote: all cars can face either front or back. question says one direction but does not specify a direction Then you haven't read properly. This is from the question itself: "There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction"



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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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18 May 2019, 09:16
The direction should remain the same. Let's that 5 parking spaces are bound by a wall on one side and open on the other. Now the bumper of all 5 cars can face the wall or the back of all 5 cars can face the wall. For every permutation, you have 2 options. Say rrrby is one order, this can be front rrrby or back rrrby. In both cases notice that all cars are facing the same direction
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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02 Sep 2019, 08:36
in this question order is matter ..... 5*5=25 is total but not answer ....you can easily choose less than 25 hence 20 !
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Re: There are 5 cars to be displayed in 5 parking spaces with all the cars
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