A group of 6 children are choosing colors of pencils to use to draw a picture. The available colors are black, red, and blue. If the second child refuses to use red pencils and the third child refuses to use blue pencils, then how many ways are there for the children to choose pencils? Assume that there are 12 pencils available of each color and different children are allowed to choose the same color of pencils.
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To find the number of ways the children can choose pencils, we need to consider the restrictions given by the second and third child.
The first child has no restrictions and can choose any of the three colors (black, red, or blue). So, there are 3 choices for the first child.
The second child refuses to use red pencils. Therefore, this child has only two choices - either black or blue.
The third child refuses to use blue pencils. Similarly, this child also has two choices - either black or red.
Since the choices of each child are independent of one another, we can multiply the number of choices for each child:
Number of choices for Child 1 = 3
Number of choices for Child 2 = 2
Number of choices for Child 3 = 2
To find the total number of ways the children can choose pencils, we multiply these numbers together:
Total ways = Number of choices for Child 1 * Number of choices for Child 2 * Number of choices for Child 3 = 3 * 2 * 2 = 12
Therefore, there are 12 ways for the children to choose pencils.