A group of 6 children are choosing colored pencils to draw a picture. Each child is allowed to select one color. The available colors are green, 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.
A) 324
B) 36
C) 729
D) 243

Question

Quizplus · Accepted Answer

There are 12 ways for the first child to choose a pencil, 11 ways for the second child to choose a pencil (since they cannot choose red), 12 ways for the third child to choose a pencil (since they cannot choose blue), 12 ways for the fourth child to choose a pencil, 12 ways for the fifth child to choose a pencil, and 12 ways for the sixth child to choose a pencil. Therefore, there are 12 * 11 * 12 * 12 * 12 * 12 = 324 ways for the children to choose pencils.

A Group of 6 Children Are Choosing Colored Pencils to Draw