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Difficulty: MediumCounting with Restrictions and Repetitions

An interior designer is arranging a row of 7 decorative wall tiles consisting of 3 identical blue tiles, 2 identical yellow tiles, and 2 identical red tiles. If the 2 red tiles cannot be placed next to each other, how many distinct arrangements of the 7 tiles are possible?

Answer: 150

Answer

The total number of distinct arrangements possible is 150.
To find the total number of distinct arrangements where no two red tiles are adjacent, first calculate the arrangements of the 5 non-restricted tiles (3 blue, 2 yellow), which is 5!3!2!=10\frac{5!}{3!2!} = 10. Placing 5 tiles creates 6 available spaces (including the two ends). Selecting 2 of these 6 spaces for the 2 identical red tiles yields (62)=15\binom{6}{2} = 15 choices. By the Fundamental Counting Principle, the total number of valid arrangements is 10×15=15010 \times 15 = 150.

Step-by-Step Solution

1
Calculate the number of distinct ways to arrange the non-restricted tiles (3 identical blue and 2 identical yellow).
The number of distinct arrangements of the 5 non-red tiles is 5!3!2!=1206×2=10\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10.
Arranging all non-restricted tiles first creates the specific positions into which the restricted tiles can be inserted.
2
Determine the number of valid positions for the 2 identical red tiles such that no two are adjacent.
Placing 5 tiles creates 6 available insertion slots (one at each end and four between tiles). Choosing 2 distinct slots out of 6 gives (62)=6×52=15\binom{6}{2} = \frac{6 \times 5}{2} = 15 ways.
Selecting 2 distinct slots ensures that every selected space holds at most one red tile, guaranteeing that no two red tiles are adjacent.
3
Multiply the results from Step 1 and Step 2 using the Fundamental Counting Principle.
10×15=15010 \times 15 = 150.
Each arrangement of non-red tiles can be independently paired with any valid placement of the red tiles.

Key Concept

Counting arrangements of identical items with non-adjacency restrictions using the slotting method.
Estimated Time:1m 30s
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Topics

Counting with Restrictions and Repetitions

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