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Bunuel
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A researcher plans to identify each participant in a certain medical 04 Sep 2024, 02:58
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lnyngayan
Bunuel
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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 number of single-letter codes possible would be \(n\) itself;
    The number of two-letter codes possible (pair of distinct letters) 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.

Or else one could just test numbers.

    Four letters are enough for \(4+C^2_4=4+6=10\) codes. Not enough.
    Five letters are enough for \(5+C^2_5=5+10=15\) codes. Good!

Hope it's clear.

Hope it's clear.
­
Could you explain why

\(C^2_n+n\geq{12}\) becomes \(\frac{n(n-1)}{2}+n\geq{12}\)?
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Sure. This is expalined here:

https://gmatclub.com/forum/a-researcher ... l#p1344122

But here it is once again:

\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.­
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A researcher plans to identify each participant in a certain medical 04 Sep 2025, 20:55
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