Solve this problem about paintings that involves permutations

[chwala]

Homework Statement

A music curator is arranging 14 rare paintings for an exhibition.
The paintings consist of:

4 portraits,
5 landscapes,
3 abstract paintings,
and 2 modern art pieces.

All paintings are distinct. The paintings are arranged in a row along a gallery wall. Find the number of possible arrangements if all 3 abstract paintings are together, but
no portrait is adjacent to a modern painting.

Relevant Equations

Permutation.

Ok, in my thinking i treat the $3$ abstract paintings as one block and considering the 5 landscapes and 4 portraits, i now have a total of 10 distinct items; but before i proceed i was asking can we solve this by excluding the portraits and modern first? work on that Then, can we consider the modern and portraits as one block? and work on subtracting that? my question is basically on the approach; exclusive or inclusive of the modern and portraits, i seek direction on this...

 

[PeroK]

Are "modern" and "abstract" paintings the same thing. Can a portrait be adjacent to a modern art piece?

 

[chwala]

No they are not the same and neither can a portrait be adjacent to a modern art.

 

[PeroK]

No they are not the same and neither can they be adjacent to each other.

The question makes no sense. There are no modern paintings in the exhibition.

 

[PeroK]

No they are not the same and neither can they be adjacent to each other.

This answer makes no sense to me. Check your transcription of the question.

 

[chwala]

The question makes no sense. There are no modern paintings in the exhibition.

2 modern art pieces.

 

[PeroK]

2 modern art pieces.

The rule is that no portrait is adjacent to a modern art piece?

 

[PeroK]

Looks difficult! I can't immediately see a shortcut. Focus on the two modern art pieces, perhaps?

 

[PeroK]

The answer I got was $(224 \times 9!) \times 6$.

Given that the total number of permutations is $12!\times 6$, that gives the probability that no portrait is next to a modern art piece of about 0.17. Given a random permutation.

Which seems about right.

 

[PeroK]

Here's how I did it. Temporarily take the two modern art pieces as indistinguishable. We can multiply by 2 at the end.

There are 66 different possibilities for their position. These must be handled separately depending on:

Whether one or both are at the ends.
How far apart they are: next to each other; one piece between them; or, two or more pieces between them.

The 66 possibilities fall into 7 categories. I calculated the number in each category and the permutations in each category.

Hopefully I got the right answer.

 

[PeroK]

PS the key factor is how many spaces are adjacent to the modern art pieces. If they are together at the end, there is only one space (where no portrait may go). If neither is at the end and they have more than one space between them, there are four spaces where no portrait may go.

 

[chwala]

... noted @PeroK . Let me look at it with your points in mind.

 

[pbuk]

The answer I got was $(224 \times 9!) \times 6$.

Hopefully I got the right answer.

Yes, 487,710,720 verified by computation.

 

[DaveC426913]

Does nobody care why a music curator is trying to hang these rare paintings?

 

[PeroK]

... noted @PeroK . Let me look at it with your points in mind.

Identifying the 7 categories first is a good start.

 

[anuttarasammyak]

Homework Statement: A music curator is arranging 14 rare paintings for an exhibition.
The paintings consist of:

4 portraits,
5 landscapes,
3 abstract paintings,
and 2 modern art pieces.

All paintings are distinct. The paintings are arranged in a row along a gallery wall. Find the number of possible arrangements if all 3 abstract paintings are together, but
no portrait is adjacent to a modern painting.
Relevant Equations: Permutation.

Ok, in my thinking i treat the 3 abstract paintings as one block and considering the 5 landscapes and 4 portraits, i now have a total of 10 distinct items;

After the idea the problem is reduced to 12 items, i.e.,
4 portraits,
5 landscapes,
1 abstract paintings group,
and 2 modern art pieces. Permutation in abstract painting group 3! should be maltiplied to the answer or the reduced problem.

For the reduced problem we focus on positons of 2 modern art pieces.
Say we walk and see x pieces before a modern art, y pieces between the modern arts and z pieces after the next modernart to the end.
The number of positons allowed for portraits is
$$N(x,y,z)=(x-1)\theta(x-1)+(y-2)\theta(y-2)+(z-1)\theta(z-1)$$
where $\theta(x)=1$ for x>0, otherwise 0.
Thus the number of arrangements is
$$2!4!\sum_{x,y,z \in\{0,1,2,3,4,5,6,7,8,9,10\},x+y+z=10\ } \ _{N(x,y,z)}C_{4}$$
Further 6! should be multiplied for remaining 6 positions.

 

[chwala]

Amended by including $6$

We consider two cases - what i would call gap method.

Case 1 - Two modern are adjacent ... and considering 5L,2M,1A( 3 abstracts)

$6×2! ×7! ×4!× \binom{9}{4}$

Case 2 - Two modern are separated ... and considering 5L,2M,1A( 3 abstracts)

$6×(8!-(2! ×7!))×4!×\binom {8}{4}$

Case 1 + Case 2 = 487,710, 720.

 

[anuttarasammyak]

After the idea the problem is reduced to 12 items, i.e.,
4 portraits,
5 landscapes,
1 abstract paintings group,
and 2 modern art pieces. Permutation in abstract painting group 3! should be maltiplied to the answer or the reduced problem.

For the reduced problem we focus on positons of 2 modern art pieces.
Say we walk and see x pieces before a modern art, y pieces between the modern arts and z pieces after the next modernart to the end.
The number of positons allowed for portraits is
$$N(x,y,z)=(x-1)\theta(x-1)+(y-2)\theta(y-2)+(z-1)\theta(z-1)$$
where $\theta(x)=1$ for x>0, otherwise 0.
Thus the number of arrangements is
$$2!4!\sum_{x,y,z \in\{0,1,2,3,4,5,6,7,8,9,10\},x+y+z=10\ } \ _{N(x,y,z)}C_{4}$$
Further 6! should be multiplied for remaining 6 positions.

$$N(x,y,z)=\max(x-1,0)+\max(y-2,0)+\max(z-1,0)$$
$$=\max(x-1,0)-x+\max(y-2,0)-y+\max(z-1,0)-z+10$$
$$=10-\min(x,1)-\min(y,2)-\min(z,1)$$
Thus N={6,7,8,9}

Coefficient for N=6: 6 items are placed Left, Between and Right of " XMX XMX "
Number of ways is same as that of 6 be expressed as sum of three not negative digits
6=0+0+6=0+1+5=.. 32 ways

Coefficient for N=7: 7 items are placed Left and Between of " XMX XM"
Number of ways is same as that of 7 be expressed as sum of two not negative digits
7=0+7=1+6=.. 8 ways
With right-left turn 8 ways
7 items are placed Right and Left of " XMXMX " 8 ways
Thus 24 ways

Coefficient for N=8: 8 items are placed Between of"MX XM" 1 way
8 items are placed Right and Left of " XMMX " 9ways
Thus 10 ways

Coefficient for N=9: 9 items are placed Right of "MMX " 1 way
Toghether with right-left turn thus 2 ways

Total 32+24+10+2=66 ways.

 

[pbuk]

Case 1 + Case 2 = 487,710, 720.

You forgot to multiply by 6 (Case 1 + Case 2 = 81,285,120).

That is an elegant solution - was it inspired by @PeroK's (or @anuttarasammyak's frankly unfathomable) approach?

 

[PeroK]

We consider two cases - what i would call gap method.

Case 1 - Two modern are adjacent ... and considering 5L,2M,1A( 3 abstracts)

$2! ×7! ×4!× \binom{9}{4}$

Case 2 - Two modern are separated ... and considering 5L,2M,1A( 3 abstracts)

$(8!-(2! ×7!))×4!×\binom {8}{4}$

Case 1 + Case 2 = 487,710, 720.

Can you explain how those calculations work?

 

[chwala]

You forgot to multiply by 6 (Case 1 + Case 2 = 81,285,120).

That is an elegant solution - was it inspired by @PeroK's (or @anuttarasammyak's frankly unfathomable) approach?

Will fix that...

 

[chwala]

Can you explain how those calculations work?

Case 1

Treat the two modern paintings as one block, leaving 7 units to arrange: 5 landscapes, 1 modern block, and 1 abstract block. Arrange these in 7! ways, with 2! arrangements within the modern block and 3! arrangements of the abstracts. Finally, arrange the 4 distinct portraits in 4! ways and select 4 of the 9 available gaps using $\binom {9}{4}$

Case 2
First, arrange the 8 units (5 landscapes, 2 modern paintings, and 1 abstract block) in 8! ways. Subtract the arrangements where the modern paintings are adjacent, 2! × 7!, leaving only arrangements where they are separated. Then account for the internal abstract arrangements, the 4 distinct portraits and their placement in the selected gaps.