Alternate method

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Alternate method

by knight247 » Wed Sep 21, 2011 4:49 am
A company that ships boxes to a total of 12 distribution centres uses colour coding to identify each centre. If either a single colour or a pair of two different colours is chosen to represent each centre and if each cenre is uniquely represented by that choice of one or two colours, what is the minimum number of colours needed for that coding, assuming that the order of the colours in a pair do not matter?
(A)4
(B)5
(C)6
(D)12
(E)24

OA is B

The method I followed was to look at the answer choices directly i.e. (A) 4 colours will cover 4 distribution centres and 4C2=6 so 6 more centres will be covered. 6+4=10 so two centres will still be left so incorrect.

(B)5 colours will cover 5 centres. Then 5C2=10 and 5+10=15 so this will cover all the twelve centres with a few combinations left over. So B

However, I'm hoping to figure out a direct method to solve this one. Hoping to get responses on an alternate method that is more direct. Detailed explanations would be highly appreciated.
Source: — Problem Solving |

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by shankar.ashwin » Wed Sep 21, 2011 5:02 am
Your method is the fastest I could think of, but an alternate approach to it would be;

You could use single or combination of 2 colors to represent a centre;

Let n be the minimum No, of colors required, So

nC2 + n > 12 (N for single color codes and nC2 for 2 color combinations)

n(n-1)/2 + n >12

SImplifying; n^2+n >24

n(n+1) > 24 (consecutive integers)

So considering only integer values; minimum value n can take is 5.

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by cans » Wed Sep 21, 2011 5:04 am
using options is better. But here goes the algebraic solution:
let min colors be n.
then n + nc2 >=12 or n(n+1)>24
if n=4, 20<24
n=5, 30>24
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by knight247 » Wed Sep 21, 2011 5:06 am
Great stuff Ashwin...Thanks Bro