swerve 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
The OA is B.
Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.
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
