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A researcher plans to identify each participant in a certain medical

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New post 06 Nov 2018, 10:11
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 28 Jan 2019, 08:50
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Got tripped up a bit on the wording, but saw the familiarity of the problem. The hard part is that these combination/permutation word problems express things in an particular way that doesn't always make it immediately clear whether order matters or not.

The issue I had was with the use of the word order, which in my mind would typically be associated with permutations, but the restriction of the problem actually means it's a combination.

"written in alphabetical order" - meaning that the letters have to be alphabetically ordered, but not necessarily consecutively next to each other on the alphabet. This actually makes it a combination and not a permutation question, because it removes all the other possibilities (i.e. we can have AB but not BA). Drawing it out makes sense:

Image

The fact that the letters have to be ordered alphabetically is a convoluted way of saying only unique combinations of letters matter. Hopefully my thinking is correct in this regard.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 28 Jan 2019, 20:17
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


Constraints:
"Single Letter or a pair of distinct letters written in alphabetical order"

List the possibilities out:
A, AB, B, AC, BC, C, AD, BD, CD, D, AE, BE <= 12th combo

A, B, C, D, E were used, 5 different letters.

Answer is B
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 02 Apr 2019, 08:13
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


Bunuel, chetan2u

Can any one elaborate condition of highlighted part.
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New post 02 Apr 2019, 09:05
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Gmatprep550 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


Bunuel, chetan2u

Can any one elaborate condition of highlighted part.


It means that the letters we choose must be presented in alphabetical order.
So, for example, EG is good, since E comes before G in the alphabet.
Conversely, GE is not acceptable, since G does not come before E in the alphabet.

Cheers,
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A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 20 Apr 2019, 19:47
A simpler way would be to see the first option, which says 4.
If we take 4 alphabets then the number of words with
one alphabet : 4
two alphabets(considering 4 & sequence) : 3
two alphabets(considering 3 & sequence) : 2
two alphabets(considering 2 & sequence) :1
Total =10
Hence the next higher to 4 ie 5 is the correct option
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 29 May 2019, 01:02
Bunuel, can you explain why this is a combination and not a permutation?


Bunuel wrote:
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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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 29 May 2019, 19:32
Hi crushorange,

The last sentence in the prompt tells us: "Assume that changing the color order within a pair does NOT produce different codes." This means that two different colors will only produce ONE code. For example Blue-Green and Green-Blue are the SAME code. Thus, the order of the colors does NOT matter and we're dealing with a Combination.

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New post 12 Jul 2019, 10:20
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A,B,AB,C,AC,BC,D,AD,BD,CD,E,AE....

5 Letters and we can get 12 code in alphabetical order.

Is it the right way of doing it?
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New post 12 Jul 2019, 11:43
Hi sk710,

Based on the work that you've shown, you used a 'brute force' approach (re: you just 'mapped out' the possibilities without the need of a formula or any excessive 'math'). That's actually a great way to approach this question - and you'll find that a certain number of Quant questions (including many DS questions) can be solved by taking that same general approach.

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New post 14 Sep 2019, 14:50
I recommend do not go for the Formula, because it may confuse some people.

Follow the Basic Logic, and solve it in 1 Min.

Ans. Requirement -
1.) It should be a SINGLE letter or a PAIR.
2.) The code should be in Alphabetical Order.

So, let's start a Quick count -
A - Letters Used = 1
B - Letters Used = 2
AB - Letters Used = 2
C - Letters Used = 3
AC - Letters Used = 3
BC - Letters Used = 3
D - Letters Used = 4
AD - Letters Used = 4
BD - Letters Used = 4
CD - Letters Used = 4
E - Letters Used = 5
AE - Letters Used = 5

12 Participants have been Covered and Letter Used Till Now is 5 i.e. (A, B, C, D, E)

Forget the Formulas in Quant, just go for the Logic, think how you could solve any question if it were a real-life Problem :)
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 18 Sep 2019, 09:42
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


With 4 letters say A B C D
A
B
C
D
AB
AC
AD
BC
BD
CD
10 Possible codes.

With 5 you will definitely have more, but just to list

A
B
C
D
E
AB
AC
AD
AE
BC
BD
BE and so on..
So 6 will be sufficient for 12 participants.
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New post 10 Oct 2019, 21:13
Bunuel wrote:
eaakbari wrote:
Bunuel wrote:

Practice: try to use the same concept.


Okay here goes,

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)

Thus

\(nC3 + nC2 + n\)> \(12\)

\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)> \(12\)

Simplifying

\(n*(n^2 +5)\)> \(72\)

Only sufficient value of \(n = 4\)

Is it correct?


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.

Bunuel plz claer this :
i though alphabetial menas AB,BC ,CD only
AC ,AD also alphabetical?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 10 Oct 2019, 21:18
vanam52923 wrote:
Bunuel wrote:
eaakbari wrote:
Okay here goes,

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)

Thus

\(nC3 + nC2 + n\)> \(12\)

\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)> \(12\)

Simplifying

\(n*(n^2 +5)\)> \(72\)

Only sufficient value of \(n = 4\)

Is it correct?


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.

Bunuel plz claer this :
i though alphabetial menas AB,BC ,CD only
AC ,AD also alphabetical?


I think my post answers your question.
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Re: A researcher plans to identify each participant in a certain medical   [#permalink] 10 Oct 2019, 21:18

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