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.
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Almost identical question:

John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
A. 24
B. 12
C. 7
D. 6
E. 5

The concept is not that hard. We can use combination or trial and error approach.

Combination approach:
Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Trial and error approach:
If the minimum number of colors needed is 4 then there are 4 single color codes possible PLUS \(C^2_4=6\) two-color codes --> 4+6=10<12 --> not enough for 12 codes;

If the minimum number of colors needed is 5 then there are 5 single color codes possible PLUS \(C^2_5=10\) two-color codes --> 5+10=15>12 --> more than enough for 12 codes.

Actually as the least answer choice is 5 then if you tried it first you'd get the correct answer right away.

Answer: E.

Hope it helps.
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The number of letters should be enough to make at least 12 codes, thus the number of codes must be more than or equal to 12.
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Notice that we are told that letters in the code should be written in alphabetical order. Now, 2Cn gives different pairs of 2 letters possible out of n letters, but since codes should be written in one particular order (alphabetical), then for each pair there will be only one ordering possible, thus the number of codes out of n letters equals to number of pairs out of n letters.

Hope it's clear.
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Practice: try to use the same concept.
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Correct.

Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.

Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
AD;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.
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Hi Bunuel,

From n letters we choose the number of pairs, the result will be \(C^2_n\) which may include 2 kinds of pairs (AB) and (BA). Still confused .

we can take 1,2 and 3
like
A, B, C
AB, BC
ABC

Why did you ignored possibility of 3 or 4 alphabets taken together, this will give us 4 letters?
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Lets take A B C D
A
B
C
D
AB
AC
AD
BC
BD
CD
ABC
BCA
CBA

It is alphabetical and all letter for a particular codes are different.
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I have a questions here:
How did we get from \(n(n+1)\geq{24}\) to \(n_{min}=5\)