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
Medical experiment
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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
The least n with (n^2)/2>12 is n=5.
Antoni
- PussInBoots
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It should be N {combinations of 1 letter code} + N * (N-1) / 2 {combinations of 2 letter code} >12
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Can someone tell me which topic is the question testing? Formulas looks foriegn to me. Pls help.
- PussInBoots
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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.
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.