Bunuel wrote:
A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)
(A) 4
(B) 5
(C) 6
(D) 12
(E) 24
Problem Solving
Question: 132
Category: Arithmetic Elementary combinatorics
Page: 79
Difficulty: 600
The statement says
a single color or a pair of two different colors is chosen to represent each center.
Hence, the number of combinations have to be greater or equal than 12:
\(nC1 + nC2 >= 12\)
Where:
\(nC1 = n\)
\(nC2 = \frac{n*(n-1)}{2}\)
So:
\(nC1 + nC2 = n+\frac{n*(n-1)}{2} >= 12\)
\(\frac{2n+n*(n-1)}{2} >= 12\)
\(2n+n*(n-1) >= 24\)
\(n2+n >= 24\)
We can now pick values for n, which will be faster than solving:
If n=4,
\(n2+n = 20 < 24\)
If n=5,
\(n2+n = 29 >= 24\)
Answer: B (n=5)For those willing to solve for n:
\(n2+n >= 24\)
\(n2+n -24 >0\)
Solving for n,
\(\frac{-1+-sqroot(1+96)}{2}\)
\(\frac{-1+-sqroot(97)}{2}\)
We don't really need to solve the square root:
- Negative option of the square root is not possible
- The positive option can be approximated by
\(\frac{1+sqroot(100)}{2} = (approx.)= 5\)
(but slightly less than 5)
Given that the inequality is:
\(n2+n -24 >0\)
Any value of n>=(slightly less than 5) will make the inequality positive.
Hence
n=5Answer: B (n=5)