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Seven cars of seven different models are going to park in a row of

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Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 17 Mar 2017, 07:46
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Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

a) 600
b) 720
c) 1440
d) 4320
e) 4800

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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 17 Mar 2017, 07:55
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daboo343 wrote:
Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

a) 600
b) 720
c) 1440
d) 4320
e) 4800


Consider models P and Q as one unit - {PQ}. Now, we'll have 6 units: {1}{2}{3}{4}{S}{PQ}. These 6 units can be arranged in 6! ways. P and Q within their unit can be arranged in 2 ways: PQ and QP. Thus the total ways to arrange 6 units is 6!*2.

Finally, in half of these arrangement S will be somewhere to the right of {PQ} and in the remaining half of these arrangements S will be somewhere to the left of {PQ} (how else? why should the number of arrangements favour one over another?).

Therefore, the answer is 6!*2/2 = 720.

Answer: B.
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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 17 Mar 2017, 07:57
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Bunuel wrote:
daboo343 wrote:
Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

a) 600
b) 720
c) 1440
d) 4320
e) 4800


Consider models P and Q as one unit - {PQ}. Now, we'll have 6 units: {1}{2}{3}{4}{S}{PQ}. These 6 units can be arranged in 6! ways. P and Q within their unit can be arranged in 2 ways: PQ and QP. Thus the total ways to arrange 6 units is 6!*2.

Finally, in half of these arrangement S will be somewhere to the right of {PQ} and in the remaining half of these arrangements S will be somewhere to the left of {PQ} (how else? why should the number of arrangements favour one over another?).

Therefore, the answer is 6!*2/2 = 720.

Answer: B.


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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 07 Aug 2018, 15:48
I really didn't understand why we have to divide by 2 when we consider the "S" to the right or left.
How can we get in this magic number?
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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 08 Aug 2018, 08:19
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pamelaover wrote:
I really didn't understand why we have to divide by 2 when we consider the "S" to the right or left.
How can we get in this magic number?


pamelaover

Let's say we have 3 letters, A, B ad C.

What are the different arrangements possible? 6 - ABC, ACB, BCA, BAC, CAB, CBA. Now, if we are asked, in how many such arrangements are C right of B? It is 3 - ABC, BCA, BAC, half of the total number of 6. So, when we are asked, in how many such arrangements, S is to the right of P, we can get that number by dividing the total number of arrangements by 2.
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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 12 Aug 2018, 04:02
Lets say our cars are A, B, C, D, P, Q, and S.

P-Q-A-B-C-D-S

If we keep S ant the last (7th) spot, then we have 5!2 number of ways to arrange the cars, so that P and Q will be next to each other. If we take P and Q as one element then PQ-A-B-C-D could be arranged in 5! ways. Because if we replace place P with Q then we will have 5!2 number of arrangements.

P-Q-A-B-C-S-D

Keeping S in the 6th spot.

We could arrange cars to the left of S in 4!2 ways. Because, taking P and Q as one element PQ-A-B-C could be ordered in 4! ways. Changing place of P and Q will double the number of arrangements, so we will have 4!2 different arrangements of cars to the left of S. Because of the fact that we can place 4 different cars in the 7th place (which is to the right of car S) the seven cars could be arranged in 4!2*4 ways.

P-Q-A-B-S-C-D

S in the 5th spot

Cars to the left of S could be arranged in 3!2 ways. Cars to the right of S could be arranged in 4*3 ways. Therefore cars could be arranged in 3!2*4*3 ways.

P-Q-A-S-B-C-D

S in the fourth spot. We can arrange cars to the left of S in 2!2 ways. Cars to the right of S can be arranged in 3*2 ways. Therefore in this scenario we have total of 2!2*3*2 different arrangements of cars.

P-Q-S-A-B-C-D

S in the 3rd spot

Cars to the left S could be arranged in just 2 ways. Cars to the right of S could be arranged in 4*3*2 ways. Therefore total number of arrangements in this scenario is 2*4!

Therefore answer is 5!2+4!2*4+3!2*12+4!*2*2+2*4! = 720
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Re: Seven cars of seven different models are going to park in a row of  [#permalink]

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New post 01 Feb 2019, 17:37
daboo343 wrote:
Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

a) 600
b) 720
c) 1440
d) 4320
e) 4800


If models P and Q must park next to each other, we can consider them as one unit. That is, there are 6 cars (instead of 7) and they can be arranged in 6! = 720 ways. However, since P and Q are considered one unit, they can be arranged as either PQ or QP. Therefore, there are 720 x 2 = 1440 ways to arrange these cars if there are no other restrictions.

However, we have a restriction that model S must be on the right of P and Q. Model S is to the right of P and Q in exactly half of the 1440 ways, and it is to the left of P and Q in the remaining half of the 1440 ways. Therefore, the number of ways S will be to the right of P and Q is ½ x 1440 = 720.

Answer: B
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Re: Seven cars of seven different models are going to park in a row of   [#permalink] 01 Feb 2019, 17:37
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