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# A researcher plans to identify each participant in a certain

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Manager
Joined: 07 Jun 2006
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A researcher plans to identify each participant in a certain [#permalink]

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17 Nov 2006, 10:07
3
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45% (medium)

Question Stats:

59% (00:54) correct 41% (01:16) wrong based on 190 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Dec 2012, 22:52, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.

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Director
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17 Nov 2006, 10:51
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

I think the order matters so I'm going w/B. I reasoned this one out.
Does "a pair of distinct letters written in alphabetical order" mean anything significant?

1 2 3 4 5 6 7 8 9 10 11 12
A B C D E AA AB AC AD AE BA BB

Last edited by ggarr on 17 Nov 2006, 11:26, edited 3 times in total.

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Senior Manager
Joined: 08 Jun 2006
Posts: 335

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Location: Washington DC

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17 Nov 2006, 11:13
Getting 5

Bcos 5C2 + 5 = 15
using 4 we can reach up to 4C2 + 4 = 10

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Manager
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17 Nov 2006, 11:15
5
1 2 3 4 5 6 7 8 9 10 11 12
A B AB C AC BC D AD BD CD E AE

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Senior Manager
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17 Nov 2006, 13:30
B - 5

4C2 + 4 = 10 Not enough

5C2 + 5 = 15 Enough

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Intern
Joined: 14 Nov 2006
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17 Nov 2006, 14:03
Sorry guys, but how do you calculate 4C2 or 5C2?

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VP
Joined: 25 Jun 2006
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19 Nov 2006, 17:40
B 2.

the equation you get is:

x + xC2 >= 12.

then plug in from the smallest number.

order does not matter because it has to be ordered in alphabetical order, so it is just like combination instead of permutation.

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19 Nov 2006, 18:52
A
B
AB
C
BC
AC
D
BD
CD
E

So 5 letters needed

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Manager
Joined: 10 Sep 2012
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15 Dec 2012, 09:22
ywilfred wrote:
A
B
AB
C
BC
AC
D
BD
CD
E

So 5 letters needed

This question is pissing me off because its not designed well...

"or a pair of distinct letters written in alphabetical order."

That clearly means that AD, AC, cannot be used.... these are not in alphabetical order.

I got 7.

A, B, C, D, E, F, G, AB, BC, CD, DE, EF

Obviously the official answer is 5 but can someone please explain to me whe

"a pair of distinct letters written in alphabetical order" is not nullifying AC or AD.... AC is not in alphabetical order.

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Math Expert
Joined: 02 Sep 2009
Posts: 42627

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16 Dec 2012, 22:53
anon1 wrote:
ywilfred wrote:
A
B
AB
C
BC
AC
D
BD
CD
E

So 5 letters needed

This question is pissing me off because its not designed well...

"or a pair of distinct letters written in alphabetical order."

That clearly means that AD, AC, cannot be used.... these are not in alphabetical order.

I got 7.

A, B, C, D, E, F, G, AB, BC, CD, DE, EF

Obviously the official answer is 5 but can someone please explain to me whe

"a pair of distinct letters written in alphabetical order" is not nullifying AC or AD.... AC is not in alphabetical order.

AD is in alphabetical order, DA is not.
AC is in alphabetical order, CA is not.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html
_________________

Kudos [?]: 135816 [0], given: 12714

Re: Re:   [#permalink] 16 Dec 2012, 22:53
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