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Difficulty: Very hardBasic Probability and Counting Methods

An art gallery owner wants to display 5 paintings in a row on a wall. The owner selects these 5 paintings from a collection of 4 different landscape paintings and 4 different portrait paintings. The display must meet the following guidelines:

1. No two landscape paintings can be placed next to each other.
2. At least 2 landscape paintings must be displayed.

How many different arrangements of 5 paintings are possible?

Answer: 2016 arrangements

Answer

2016
The correct answer is 2016. By breaking down the problem into two mutually exclusive cases based on the number of landscape paintings (either 3 landscapes and 2 portraits, or 2 landscapes and 3 portraits), we can find the valid arrangements for each. For 3 landscapes, the only valid layout is L−P−L−P−LL-P-L-P-L, yielding P(4,3)×P(4,2)=288P(4,3) \times P(4,2) = 288 arrangements. For 2 landscapes, there are (42)=6\binom{4}{2} = 6 layout configurations, each yielding P(4,2)×P(4,3)=288P(4,2) \times P(4,3) = 288 arrangements, for a total of 1728. Adding these two cases gives 288+1728=2016288 + 1728 = 2016.

Step-by-Step Solution

1
Determine the possible number of landscape paintings (kk) that can be displayed.
k=2k = 2 or k=3k = 3
Since at least 2 landscapes must be displayed, k≥2k \ge 2. Since no two landscapes can be adjacent in a 5-painting row, we cannot have 4 landscapes (as that would require at least 3 portraits to separate them, making the total count at least 7 paintings). Thus, kk can only be 2 or 3.
2
Calculate the arrangements for the case with 3 landscape paintings and 2 portrait paintings.
288 arrangements
For 3 landscapes (LL) and 2 portraits (PP) to have no adjacent landscapes, the only possible pattern of positions is L−P−L−P−LL-P-L-P-L. The number of ways to choose and arrange 3 landscapes from 4 is P(4,3)=4×3×2=24P(4, 3) = 4 \times 3 \times 2 = 24. The number of ways to choose and arrange 2 portraits from 4 is P(4,2)=4×3=12P(4, 2) = 4 \times 3 = 12. The total arrangements for this case is 24×12=28824 \times 12 = 288.
3
Calculate the arrangements for the case with 2 landscape paintings and 3 portrait paintings.
1728 arrangements
For 2 landscapes (LL) and 3 portraits (PP) to have no adjacent landscapes, we place the 3 portraits first: _P_P_P_\_ P \_ P \_ P \_. We choose 2 of the 4 available spaces for the landscapes in (42)=6\binom{4}{2} = 6 ways. For each pattern, the number of ways to choose and arrange 2 landscapes from 4 is P(4,2)=12P(4, 2) = 12, and the number of ways to choose and arrange 3 portraits from 4 is P(4,3)=24P(4, 3) = 24. The total arrangements for this case is 6×12×24=17286 \times 12 \times 24 = 1728.
4
Sum the arrangements from both cases.
2016 arrangements
Since the two cases are mutually exclusive, we add their individual counts: 288+1728=2016288 + 1728 = 2016.

Key Concept

Permutations and Combinations with Constraints
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Topics

Basic Probability and Counting Methods

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