Jevan must paint 3 rooms in a house. Room A can be painted orange, red or green. Room B can be painted orange, white or red. Room C can be painted white, red or green. The 3 rooms cannot all be painted the same color. In how many different ways could Jevan paint the 3 rooms?
1) 24
2) 26
3) 27
4) 63
5) 64

2) 26

Step-by-step explanation:

The first thing to paint is to divide it into three parts, like this:

1st part:

Room A can be painted orange, red or green.

Therefore, we have 3 ways to do it.

2nd part:

Room B can be painted orange, white or red.

Therefore, we have 3 ways to do it.

3rd part:

Room C can be painted white, red or green.

Therefore, we have 3 ways to do it.

To know the total number, each part must be multiplied, that is:

3 * 3 * 3 = 27

However, some of these 27 possible results break the condition that the 3 rooms cannot be painted the same color, that only happens when the three rooms are painted red.

Therefore, the number of times would be:

27 - 1 = 26

the sum of all the angles in a triangle are equal to 180

so you can add   the 3 given angles together to equal 180:

5y + 2y + 2y = 180

or since all three angles have the same letter, combine the 3 angles to equal 180:

9y = 180

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step-by-step explanation: