rvshchwl wrote:
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.
Hi
Bunuel, I am a confused by the answer. The question stem specifically mentions that the 2-letter codes need to be alphabetized. Doesn't this mean that order *does* matter in this question, so it's a permutation, not a combination.
So instead of \(C^2_n\), wouldn't it be \(P^2_n\) --> \(n(n-1)\). We only divide by \(2!\) when we are dealing with combinations where order doesn't matter, but with permutations where order does matter, the formula is different.
Can you explain the logic here? Thanks.
Hi rvshchwl,
To start, this question can be solved rather easily with a 'brute force' approach (meaning that you don't need any special formulas, you just need to 'map out' the possibilities on your pad until you have the answer to the question that is asked - and several of the posts in this thread showcase that approach). If you do choose to approach this question with a formula, then it might help to think of a simple example first; that way, you can select the formula that fits the same logic.
The prompt tells us that any 2-letter codes MUST be written in alphabetical order. For example, if we had just two letters (A and B), then we could form just ONE code:
AB
By extension, with every pair of letters, we have just ONE code. Mathematically speaking, that's a Combination, not a Permutation (with 2 different letters and a Permutation, we would have 2 codes - but again, that's not allowed here).
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