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I stuck with the following combinatorics problem: A street contains 25 houses in a row. There are 5 colours of paint available to paint the houses. The colours are red, green, blue, orange and white.

I have two question: a. In how many ways can the houses be painted if each house has to be painted using two different colours and there are no restrictions on how many times a colour can be used? (It doesn't matter how to two colours are used on the house.)

b. Each house is painted using two colours. One colour for the house and the other for the windows and doors. In how many ways can this be done.

I have come up with the following answer for question a: $({5 \choose 2})^{25}$

And for the question b I came up with the following: $(2 \cdot 10)^{25}$

I was told those answers are correct, however can someone please explain to me as detailed as possible, why those answers are this way. I am very confused with this combinatorics thing, thanks in advance.

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For (a), there are 5 choose 2 possible ways of choosing 2 colors from a set of five colors. That's then the number of ways one house can be painted. Since there are 25 houses, that is raised to the 25th power (because each particular house could be paired with any combination of the other 24), as you have done.

For (b), instead of $2 \times 10$, I would think of it as $5 \times 4$ since there are $5$ ways to pick the color of the house and then, since they can't be the same color as the house, $4$ ways to pick the colors of the doors and windows. Again, that is raised to the 25th since there are 25 houses.

Does this detail clarify it at all for you?

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  • $\begingroup$ Hey Shane, Thanks for clarifying this for me! You have let me see it from the way I thought it was initially too, and yes, it DOES make sense to see it as 5 x 4! I fully understand it now. I have accepted your answer. $\endgroup$ – Roël Gonzalez Nov 30 '14 at 18:03

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