A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a singl

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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

[spoiler]OA=B[/spoiler]

Source: Official Guide
Source: — Problem Solving |

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M7MBA wrote:
Wed May 20, 2020 7:19 am
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

[spoiler]OA=B[/spoiler]

Source: Official Guide
Since the answer choices are so small, we might consider testing cases.
So, at this point, we need to figure out where to begin. This is where we need a bit of number sense.
I chose to start with A, because I could think of several possible codes using 4 letters.

So, let's manually list all of the codes that can be made with 4 letters
We get: A, B, C, D, AB, AC, AD, BC, BD, CD
10 possible codes in total (we need at least 12 codes)

So, we need to use MORE LETTERS.
At this point, it should be obvious that adding 1 letter will give us at least 2 extra codes, so the answer is 5 letters

Answer: B

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Brent
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M7MBA wrote:
Wed May 20, 2020 7:19 am
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

[spoiler]OA=B[/spoiler]

Source: Official Guide

Solution:

We can solve this problem by using combinations. We can let n denote the total number of letters needed. Since there are 12 participants and we have the option of one-letter codes and two-letter codes, we need the sum of nC1 and nC2 to be at least 12. That is:

nC1 + nC2 ≥ 12

From here, the easiest way to solve this problem is to test the answer choices. We can start with answer choice A.

A. 4

4C1 + 4C2 = 4 + (4x3)/2 = 4 + 2x3 = 10

If we use 4 letters, we will obtain only ten unique codes; this is insufficient for 12 participants.
B. 5

5C1 + 5C2 = 5 + (5x4)/2 = 5 + 5x2 = 15

We obtain 15 unique codes by using 5 letters, and this is more than sufficient for 12 participants.

Alternate Solution:

Let’s try 4 letters. For A, B, C, D, we would have A, B, C, D, AB, AC, AD, BC, BD, and CD, which is 10 unique codes.

It is pretty obvious that since 4 letters gave us 10 codes, that 5 letters will easily create 12 or more codes. But let’s illustrate the solution, to verify: we have A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. These 15 codes are more than adequate.

Answer: B

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