OG (10th) - PS - #281

[Guest]

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. IF either a single color a pair of two different colors is chosen to represent each center and if each center is uniquely repsented by that chocie of one or two colors, what is the minimum number of colors needed for the coding? (assume that order of the colors in a pair does not matter)
A) 4
B)5
C) 6
D)12
E) 24

[GMAT 2007]

There are two appoaches two proceed: -

First approach: -

From the question, we know the color code would either be one color or combination of two. Now the total no. of distribution centers: - 12

So, if we pick three colors Red(R), Blue(B), Green (G) then possible color combinations R,B,G,RB, RG,BG.
Total possible combinations = 6

If we pick 4 colors (add Yellow to it). Possible combinations = R,B,G,Y,RB,RG,RY,BG,BY,GY

Total possible combinations = 10

If you add one more to it, would give you more then 12 combinations for sure. So (B) is the answer.

Another quicker approach to calculate combinations: -

Since we know order doesn't matter, Hence:

Total possible combinations = Individual color + combination of two colors

Now if we pick three colors

Total possible combinations = 3 + 3!/2! = 3+3 = 6

Now pick four colors

Total possible combinations = 4+ 4!/2!2! = 4+6 = 10

Now pick 5 colors

Total possible combinations = 5 + 5!/3!2! = 5 + 10 = 15 (Sufficient)

Answer is (B)

GMAT 2007

[Guest]

i still don't get how you came up with the answer B. Sorry, you went from "Total possible combinations = 5 + 5!/3!2! = 5 + 10 = 15 (Sufficient)" to assuming that the answer is B?

I just don't see it.

[StaceyKoprince]

GMAT 2007 is calculating the number of combinations possible given a certain starting point of # of colors.

I'll start with the 4 color option, since we wouldn't want to try 3 - that's not in the answer choices.

Now pick four colors

Total possible combinations = 4+ 4!/2!2! = 4+6 = 10

So we have two overall options: we can use each color individually or we can use them in combinations of two. We have four colors, so there are four ways to use them individually. That's where the first "4+" comes from.

The second option is to use them in combos of two. If I have 4 colors, how many different ways can I combine two of them? That's n! / [r!(n-r)!]. In this case, n=4 and r=2, so we have 4!/(2!2!). That's the second half of the above.

If I simplify down, they add to 10. That means that if I start with 4 colors, I have 10 possible ways to use either one color or two colors. I need to give 12 factories unique color schemes, though, so I don't have enough ways to do that.

I suspect at this point that, if 4 colors aren't quite enough, 5 will probably be enough. 10 is pretty close to 12. But I check anyway:

5 colors alone = 5 ways
5 colors in pairs of 2 = use the same formula above to get 5!/(3!2!) = 10.
Add them up to get 15 ways to use five colors either individually or in pairs. I only need to have 12 color schemes for my 12 factories, so that's enough.