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Combination or Permutation Problem

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Combination or Permutation Problem

by vongdn » Wed Oct 06, 2010 8:45 pm
A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green, yellow, or pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the type described above are possible?

ANSWER is 16.

I thought I could use a combination formula for this, but I have no idea how to proceed.
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by ramannjit » Wed Oct 06, 2010 9:31 pm
vongdn wrote:A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green, yellow, or pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the type described above are possible?

ANSWER is 16.

I thought I could use a combination formula for this, but I have no idea how to proceed.
You can go with the following approach: Same Size; same color
3 stick pads of same size and same color cane be chosen in 2c1*4c1=8 ways

Same Size; diff color
3 stick pads of same size and diff color can be chosen in 2C1 * 4C3=8 ways

8+8 = 16
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by Rahul@gurome » Wed Oct 06, 2010 9:46 pm
Let the two sizes be 1 and 2.
Let the varieties in size 1 be B1, G1, Y1, and P1 where B, G, Y and P are the four given colors, blue, green, yellow and pink.
Also let the varieties in size 2 be B2, G2, Y2 and P2.

The first kind of package has 3 notepads of same size and color.
This means all three are either all B1, or all B2 , or all G1, or all G2, or all Y1, or all Y2, or all P1 or all P2.
So we have 8 possibilities.

The second kind of package has 3 notepads of same size but different colors.
This means all 3 are either selected from B1, G1, Y1 and P1 in 4C3 = 4 ways or all 3 are selected from B2, G2, Y2 and P2 in 4C3 = 4 ways.
This means we can have 4+4 = 8 possibilities.

Adding the possibilities of both types of packages, we get 8+8 = 16 ways.

The correct answer is (C).
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