Suppose we have a horizontal stripe of length n that we wish to tile. The tiles
ID: 3184011 • Letter: S
Question
Suppose we have a horizontal stripe of length n that we wish to tile. The tiles are available in the following shapes and colors: 1x1 squares available in two colors: red and green . 1x2 rectangles available in three colors: blue, black, and yellow The figure below shows an example of a tiling for n 11 (the tiles from left to right are 1x1 red, 1x2 blue, 1x1 red, 1x2 black, 1x2 yellow, 1x2 blue, 1x1 green): a. How many different ways can we tile a stripe of length 0? How many different ways can we tile a stripe of length 17 22 3? For ful credit, make sure it is clear how you came up with your results (i.e., just giving the numeric values will not be sufficient). Note that tiles of the same shape and color are indistinguishable and the order of the colors in the stripe does matter. For example, there is only one way to have a stripe with 3 red tiles (since red tiles are not distinguished) and there are two ways to have a stripe with one red tile and one blue tile: (1) a red tile followed by a blue tile and (2) a blue tile followed by a red tile b. Find a recurrence relation for the number of ways a stripe of length n can be tiled. Make sure to justify your answer c. One technique we can use to solve a recurrence relation is described in Problems 3 and 4: write out the first few terms of the sequence, guess the solution, and then prove the solution using induction. Another technique is to use the following: Given a sequence (on] described by the recurrence relation an c101 c20n 2 where c, 62 e R and where 2.cr 2- has 2 distinct roots r, and r2. The solution to the recurrence relation is an-a rin + r2n for n-o, 1.2 where a and are constants. Solve your recurrence relation from part b using this second technique (you can use the quadratic formula e, if necessary). d. How many ways can a stripe of length 11 be tiled? (Show your work and simplify your answer as much as possible.)Explanation / Answer
Solution;-
Let there be 100 tiles to start with.
%age of green tiles = 10% of 100
=> No. of green tiles = 10
%age of blue tiles = 30% of 100
=> No. of blue tiles = 30
%age of red tiles = 60% of 100
=> No. of red tiles = 60
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Given: Of the green tiles, 60% are square, 30% are triangular, and the rest are circular.
=> No. of green square tiles = 60% of 10 = 6
=> No. of green triangular tiles = 30% of 10 = 3
=> No. of green circular tiles = 10% of 10 = 1
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Given: Of the blue tiles, 40% are square, 10% are triangular, and the rest are circular.
=> No. of blue square tiles = 40% of 30 = 12
=> No. of blue triangular tiles = 10% of 30 = 3
=> No. of blue circular tiles = 50% of 30 = 15
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Given: Of the red tiles, 20% are square, 30% are triangular, and the rest are circular.
=> No. of red square tiles = 20% of 60 = 12
=> No. of red triangular tiles = 30% of 60 = 18
=> No. of red circular tiles = 50% of 60 = 30
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To find:
a) What percent of all tiles are square? (Triangular?) (Circular?)
b) What percent of all circular tiles are green? (Blue?) (Red?)
c) What percent of all tiles are triangular blue tiles?
d) What percent of all blue tiles are triangular?
e) What percent of all triangular tiles are blue?
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a)
No. of square tiles = 6 + 12 + 12 = 30
=> %age of square tiles = 30%
No. of triangular tiles = 3 + 3 + 18 = 24
=> %age of triangular tiles = 24%
No. of circular tiles = 1 + 15 + 30 = 46
=> %age of circular tiles = 46%
b)
Total no. of circular tiles = 46
No. of green circular tiles = 1
=> %age of green circular tiles among circular tiles = 1/46 = 2.17%
No. of blue circular tiles = 15
=> %age of blue circular tiles among circular tiles = 15/46 = 32.61%
No. of red circular tiles = 30
=> %age of red circular tiles among circular tiles = 30/46 = 65.22%
c)
No. of blue triangular tiles = 3
=> %age of triangular blue tiles = 3%
d)
Total no. of blue tiles = 30
No. of blue triangular tiles = 3
=> %age of triangular blue tiles among blue tiles = 3/30 = 10%
e)
Total no. of triangular tiles = 3 + 3 + 18 = 24
Total no. of blue triangular tiles = 3
=> %age of triangular blue tiles among all triangular tiles = 3/24 = 12.5%
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