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22 Jul 2020, 05:07
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D
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Difficulty:
45%
(medium)
Question Stats:
75% (02:57) correct
25%
(03:16)
wrong
based on 369
sessions
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At an art gallery, there are 7 spaces to hang 14 paintings. The 14 paintings to choose from are: three by Cezanne, three by Monet, one by Renoir, two by Van Gogh, two by Pissarro, two by Degas, and one by Manet. The even-numbered spaces must hold either a Cezanne or a Monet. The first space must hold a Renoir. The fifth space must hold a Manet, and the remaining spaces must hold either a Van Gogh, a Pissarro, or a Degas. How many unique arrangements of paintings are possible?
A. 28
B. 72
C. 1200
D. 3600
E. 7776
A. 28
B. 72
C. 1200
D. 3600
E. 7776
22 Jul 2020, 06:52
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The first place must be a Renoir, and the fifth place has to be a Manet. Since there is one of each painting, there is only one way to arrange each of them.
The even numbered places have to be either a Cezanne or a Monet. We have 3 of each.
There are three even places (2nd, 3rd and 6th) in which to arrange these 6 paintings.
The number of arrangements = 6P3 = \(\frac{6!}{(6-3)!}\) = \(\frac{6! }{ 3!}\) = 6 * 5 * 4 = 120 ways
There are 2 places remaining, i.e the 3rd and the 7th in which we have to arrange any of the remaining 6 paintings (2 Van Goghs, 2 Pissaros and 2 Degas).
This can be done in 6P2 ways = \(\frac{6! }{ (6-2)!}\) = \(\frac{6!}{4!}\) = 6 * 5 = 30 ways
Therefore total number of arrangements = 1200 * 30 = 3600
Option D
Arun Kumar
The even numbered places have to be either a Cezanne or a Monet. We have 3 of each.
There are three even places (2nd, 3rd and 6th) in which to arrange these 6 paintings.
The number of arrangements = 6P3 = \(\frac{6!}{(6-3)!}\) = \(\frac{6! }{ 3!}\) = 6 * 5 * 4 = 120 ways
There are 2 places remaining, i.e the 3rd and the 7th in which we have to arrange any of the remaining 6 paintings (2 Van Goghs, 2 Pissaros and 2 Degas).
This can be done in 6P2 ways = \(\frac{6! }{ (6-2)!}\) = \(\frac{6!}{4!}\) = 6 * 5 = 30 ways
Therefore total number of arrangements = 1200 * 30 = 3600
Option D
Arun Kumar
22 Jul 2020, 07:16
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Refer attached image to get a pictorial representation of the problem
Let's talk about the four groups in which the problem is divided
1. The even-numbered spaces must hold either a Cezanne or a Monet
Now, since it is painting so each one of them is different from other. So, there are 6 different paintings - 3 Cezanne and 3 Monet
and we have 3 even places to fill
So, this part of the problem is reduced to
"In how Many ways can i Arrange 6 paintings in 3 positions"
This can be done in 6P3 ways = \(\frac{6!}{{(6-3!)}}\) = \(\frac{6!}{3!}\) = \(\frac{6*5*4*3!}{3!}\)
= 6*5*4 = 120 ways
2. The first space must hold a Renoir
This is straight forward as there is only one Renoir so can be done in only 1 way
3. The fifth space must hold a Manet
This is straight forward as there is only one Manet so can be done in only 1 way
4. The remaining spaces must hold either a Van Gogh, a Pissarro, or a Degas
Remaining spaces are 3 and 7.
Now, since it is painting so each one of them is different from other. So, there are 6 different paintings - 2 Van Gogh, 2 - Pissarro, 2 - Degas
and we have 2 places to fill
So, this part of the problem is reduced to
"In how Many ways can i Arrange 6 paintings in 2 positions"
This can be done in 6P2 ways = \(\frac{6!}{{(6-2!)}}\) = \(\frac{6!}{4!}\) = \(\frac{6*5*4!}{4!}\)
= 6*5 = 30 ways
So, Total number of ways = 120 * 1 * 1 * 30 = 3600
So, Answer will be D
Hope it helps!
Let's talk about the four groups in which the problem is divided
1. The even-numbered spaces must hold either a Cezanne or a Monet
Now, since it is painting so each one of them is different from other. So, there are 6 different paintings - 3 Cezanne and 3 Monet
and we have 3 even places to fill
So, this part of the problem is reduced to
"In how Many ways can i Arrange 6 paintings in 3 positions"
This can be done in 6P3 ways = \(\frac{6!}{{(6-3!)}}\) = \(\frac{6!}{3!}\) = \(\frac{6*5*4*3!}{3!}\)
= 6*5*4 = 120 ways
2. The first space must hold a Renoir
This is straight forward as there is only one Renoir so can be done in only 1 way
3. The fifth space must hold a Manet
This is straight forward as there is only one Manet so can be done in only 1 way
4. The remaining spaces must hold either a Van Gogh, a Pissarro, or a Degas
Remaining spaces are 3 and 7.
Now, since it is painting so each one of them is different from other. So, there are 6 different paintings - 2 Van Gogh, 2 - Pissarro, 2 - Degas
and we have 2 places to fill
So, this part of the problem is reduced to
"In how Many ways can i Arrange 6 paintings in 2 positions"
This can be done in 6P2 ways = \(\frac{6!}{{(6-2!)}}\) = \(\frac{6!}{4!}\) = \(\frac{6*5*4!}{4!}\)
= 6*5 = 30 ways
So, Total number of ways = 120 * 1 * 1 * 30 = 3600
So, Answer will be D
Hope it helps!
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