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Bred has to paint a wall with 8 horizontal stripes. He has enough paint only for 5 blue, 5 red and 5 white stripes. If he can use at most 2 colors, how many different ways he can paint the wall.

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Final answer:

Bred can paint the wall in 6 different ways using 8 horizontal stripes, considering a maximum of 2 colors out of blue, red, and white with enough paint for 5 stripes of each color.

Step-by-step explanation:

The question at hand involves combinatorics, which is a branch of mathematics dealing with combinations and permutations. Bred can paint a wall with 8 horizontal stripes using at most 2 colors from the options of blue, red, and white, with enough paint for 5 stripes of each color. Since he is using at most 2 colors, we need to calculate the number of combinations for each pair of colors as well as the individual colors. The possible pairs with their respective numbers of combinations are blue-red, blue-white, and red-white. For blue-red and blue-white, he can paint 5 blue stripes and 3 stripes of the other color, while for red-white, he can paint 5 red stripes and 3 white stripes.

The combinations for each pair would be:

Blue-Red: 5 blue + 3 red

Blue-White: 5 blue + 3 white

Red-White: 5 red + 3 white

Additionally, Bred can choose to use only one color. Thus, for each of the colors blue, red, and white, there will only be one way to paint the wall.

Therefore, the total different ways Bred can paint the wall are the sum of the combinations for each pair plus the individual color options:

For pair Blue-Red: 1 way

For pair Blue-White: 1 way

For pair Red-White: 1 way

Only Blue: 1 way

Only Red: 1 way

Only White: 1 way

Adding these up gives us a total of 6 different ways to paint the wall.

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