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Permutations and combinations

This topic has 1 expert reply and 5 member replies

Permutations and combinations

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Hello,

Need help for the following questions. (not sure about the accuracy of answer choices). Just want to validate my method of solving this problem.

Q. 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

Regards
MSD

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msd_2008 wrote:
Hello,

Need help for the following questions. (not sure about the accuracy of answer choices). Just want to validate my method of solving this problem.

Q. 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

Regards
MSD
5 cars - 3 red, 1 yellow & 1 blue

this can be set-up in an anagram form - RRRYB

No. of ways 5 cars can be arranged in 5 parking spaces - 5!
No. of ways cars could be repeated - 3!

Total permutations = 5!/3! = 120/6 = 20

OA?

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Yes. The OA is indeed 20 i.e A

However, I am not able to understand why did we divide 5! by 3!. Can you please explain a bit?

Also, why cant we follow the following method

5c3 + 5c1 + 5c1 = 10 + 5 +5 = 20

Regards
MSD

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msd_2008 wrote:
However, I am not able to understand why did we divide 5! by 3!. Can you please explain a bit?
Sure

This question can be easily converted into anagram question

How many ways can you arrange the word BAAAY.

5!/3!

Now the reason why 3! is divided is because they we have 3 identical A's

now if we dont divide it by 3!, it will take BA1A2A3Y different from BA2A1A3Y, which in actual terms we cannot, therefore we divide the total arrangements by the number of repetitions to avoid this confusion.

Hope this helps.

msd_2008 wrote:
5c3 + 5c1 + 5c1 = 10 + 5 +5 = 20
Regards
MSD
You cannot follow this, well to be honest I dont understand what you did. But all I can say is that this question is of permutations not combination.

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Parallel_chase,

I understood your explaination and as a matter of fact I have solved questions similar to this....such as how many ways can the letters of the word APPLE be arranged. Here we say 5!/2! because of 2 P's.
Dont know what I was thinking while approaching the question.
Anyways, thanks for your help!

Regards
MSD

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I know the OA is A but I have a problem with this answer.

The question specify that the cars must all be parked in the same direction. From my understanding that means that there are least two directions. So the final answer you get by using commutation should be multiplied by two. Wouldn't that make sense ?

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msd_2008 wrote:
Q. 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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