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Medical experiment

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by crackgmat007 » Tue Jul 28, 2009 4:09 pm
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 backsolved this problem and got the answer. But is there a easier way to solve? OA - B
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by ASS1991 » Tue Jul 28, 2009 6:31 pm
Call the number of letters n, then you have (n^2)/2 different codes. n different possibilities for the first letter, n-1 for the second plus one for the case the second letter is "empty". You divide by two to account for the alphabetical order.

The least n with (n^2)/2>12 is n=5.

Antoni
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by PussInBoots » Tue Jul 28, 2009 7:23 pm
It should be N {combinations of 1 letter code} + N * (N-1) / 2 {combinations of 2 letter code} >12
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by crackgmat007 » Tue Jul 28, 2009 8:43 pm
Can someone tell me which topic is the question testing? Formulas looks foriegn to me. Pls help.
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by PussInBoots » Tue Jul 28, 2009 11:28 pm
Combinatorics.
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by Rajani » Wed Jul 29, 2009 5:42 am
A AA - 2
B AB - 2
C AC BC - 3 (ALPHABETIC OREDR TO BE MAINTAINED)
D AD BD CD - 4
E - 1

2+ 2 +3+ 4 + 1 = 12 and alphabets used 5
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by pavan.mpv » Fri Feb 17, 2012 5:31 am
We can answer this question using common sense as well

we need 12 different codes (either 1 or 2(distinct) ones)

Let's consider the least number from the answer choice - 4
for 4, we have - 1,2,3,4 + 4c2 = 4*3/2=6 --> 4+6=10
But we need 12 different codes. So next big number after 4 is 5. :)
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