Richard has to paint a mural with seven horizontal stripes. He only has enough paint for four red stripes, four blue stripes, four white stripes, four black stripes, and four yellow stripes. If his patron wants at most two different colors in the mural, how many different ways can he paint the wall?

(A) 120
(B) 350
(C) 700
(D) 2,520
(E) 5,040

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The only possible combination for 7 horizontal stripes= 4 stripes of 1 color+ 3 strips of another color

So we need 2 colors; number of ways to select 2 color=5C2=10

Total possible arrangements= \(10*\frac{7!}{4!*3!}*2!\)=700

Since Richard needs to paint 7 stripes, he must use at least two colors since each color of paint is enough for only 4 stripes. However, since his patron wants at most two colors, he must use exactly two colors. The number of ways he can use 2 colors from 5 available colors is 5C2 = 10. For the pair of colors he chooses, he can paint 4 stripes of one color and 3 stripes of the other or vice versa. For example, if he chooses red and blue, he can paint 4 red stripes and 3 blue stripes or 3 red stripes and 4 blue stripes. Therefore, for a pair of colors, he has 2 choices to paint by the number of stripes of each color from the pair. Finally, once he has picked 2 colors and the number of stripes he’s going to paint with each color, there are 7!/(4! x 3!) = (7 x 6 x 5 x 4!)/(4! x 3 x 2) = 7 x 5 = 35 ways to paint the wall with the same 2 colors and the same number of stripes by each color. Therefore, in total, there are 10 x 2 x 35 = 700 different ways to paint the wall.

Answer: C
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