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A researcher plans to identify each participant in a certain

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2012, 04:20, edited 1 time in total.
Edited the question.
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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.
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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


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a-certain-stock-exchange-designates-each-stock-with-a-86656.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html

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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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New post 01 Dec 2012, 20:10
Bunnel. Thaks for the reply and merging similar topics. Can u please explain how >= 12 ?
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Hi Bunnel

won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....
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ronr34 wrote:
Hi Bunnel

won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....


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

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New post 23 Dec 2012, 13:22
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.


Bunuel,

What if the question didn't say 'pair'.

If 3 letter combinations were also permitted. How would you express it in Combination formula?
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eaakbari 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.


Bunuel,

What if the question didn't say 'pair'.

If 3 letter combinations were also permitted. How would you express it in Combination formula?


Practice: try to use the same concept.
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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?
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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.
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New post 24 Dec 2012, 01:53
Thanks a ton Bunuel,

Its crystal clear now.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 22 Nov 2013, 15:15
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 still have a little confuse in your formula \(C^2_n\). I am thinking this should be \(A^2_n\) because the 2-letter code must be in alphabetical order.

Hope to hear from you soon.

Thanks

Last edited by yenpham9 on 22 Nov 2013, 15:22, edited 1 time in total.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 22 Nov 2013, 15:19
yenpham9 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 Bunnel,

I still have a little confuse in your formula \(C^2_n\). I am thinking this should be \(A^2_n\) because the 2-letter code must be in alphabetical order.

Hope to hear from you soon.

Thanks


Please read the thread: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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Bunuel wrote:
yenpham9 wrote:
Bunuel wrote:
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 Bunnel,

I still have a little confuse in your formula \(C^2_n\). I am thinking this should be \(A^2_n\) because the 2-letter code must be in alphabetical order.

Hope to hear from you soon.

Thanks


Please read the thread: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091


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Hope this helps.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 22 Nov 2013, 15:37
Bunuel wrote:
ronr34 wrote:
Hi Bunnel

won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....


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.


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 :(.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 22 Nov 2013, 15:43
yenpham9 wrote:
Bunuel wrote:
ronr34 wrote:
Hi Bunnel

won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....


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.


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 :(.


Maybe the following example would help. Consider 4 letters {a, b, c, d}. How many 2-letter words in alphabetical order are possible? The answer is \(C^2_4=6\):
ab;
ac;
ad;
bc;
bd;
cd.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 22 Nov 2013, 15:53
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 :(.[/quote]

Maybe the following example would help. Consider 4 letters {a, b, c, d}. How many 2-letter words in alphabetical order are possible? The answer is \(C^2_4=6\):
ab;
ac;
ad;
bc;
bd;
cd.[/quote][/quote]

Thanks a lot Bunuel. I got it now :). Have a nice weekend!
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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 25 Nov 2013, 04:38
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.


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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Re: A researcher plans to identify each participant in a certain   [#permalink] 25 Nov 2013, 04:38

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