A researcher plans to identify each participant in a certain medical

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

Bunuel, chetan2u

Can any one elaborate condition of highlighted part.

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.

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.

One approach is to add a BLANK to the letters in order to account for the possibility of using just one letter for a code.

ASIDE: Notice that, if we select 2 characters, there's only 1 possible code that can be created. The reason for this is that the 2 characters must be in ALPHABETICAL order. Or, in the case that a letter and a blank are selected, there's only one possible code as well.

Now we'll test the answer choices.

Answer choice A (4 letters)
Let the letters be A, B, C, D
We'll add a "-" to represent a BLANK.
So, we must choose 2 characters from {A, B, C, D, -}
In how many ways can we select 2 characters?
We can use combinations to answer this. There are 5 characters, and we must select 2. This can be accomplished in 5C2 ways (= 10 ways).
So, there are only 10 possible codes if we use 4 letters. We want at least 12 codes.

[i]ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head below.

Answer choice B (5 letters)
Let the letters be A, B, C, D, E
Once again, we'll add a "-" to represent a BLANK.
So, we must choose 2 characters from {A, B, C, D, E, -}
There are 6 characters, and we must select 2. This can be accomplished in 6C2 ways (= 15 ways...PERFECT).

So, the least number of characters needed is 5

Answer: B

Hey,

I have a question regarding this. I decided not to use the regular formula and do it by trying. So my codes would be:

ZZ
YY
YZ
XX
XY
XZ
WW
WX
WY
WZ
Z
Y

I got 12 codes, all different, with only four letters. Can you please tell me what am I misinterpreting?

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

Bunuel , VeritasKarishma , does the word pair have no significance here? i.e. can we even take a combination like ABC,BCD,CDE? Please advise.

Bunuel , VeritasKarishma , does the word pair have no significance here? i.e. can we even take a combination like ABC,BCD,CDE? Please advise.

Can someone explain why the AB and BA are not distinct codes, ie in my view patient AB is different from patient BA.

Because with this line of logic, answer should be A.

Unless I am missing something or the whole coding industry is wrong.