Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4
B. 5
C. 6
D. 7
E. 8
Say there are minimum of \(n\) letters needed, then;
The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);
We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).
Answer: B.
Hope it's clear.
Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question.
When we say
\(C^2_n+n\geq{12}\) that means that a combination of 2 letters out of a group of n letters should yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields -1 and 0. Why am I so off here?
Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order
DOES matter? How does that affect the above equation.
P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful?
The first advice would be, and I cannot stress this enough, to read the whole thread and follow the links to similar problems.
.
or 2 letters. Now, if we have n letters how many codes we can make?
The # of pair of distinct letters codes possible would be \(C^2_n\).
So, out of n letters we can make \(n+C^2_n\) codes: n one-letter codes and \(C^2_n\) two-letter codes.
If n=5, then n(n+1)=30>24.
Hence, \(n_{min}=5\).
Notice that we have \(n(n+1)\geq{24}\) NOT \(n(n+1)\geq{0}\).