A researcher plans to identify each participant in a certain medical

Hi rvshchwl,

To start, this question can be solved rather easily with a 'brute force' approach (meaning that you don't need any special formulas, you just need to 'map out' the possibilities on your pad until you have the answer to the question that is asked - and several of the posts in this thread showcase that approach). If you do choose to approach this question with a formula, then it might help to think of a simple example first; that way, you can select the formula that fits the same logic.

The prompt tells us that any 2-letter codes MUST be written in alphabetical order. For example, if we had just two letters (A and B), then we could form just ONE code:

AB

By extension, with every pair of letters, we have just ONE code. Mathematically speaking, that's a Combination, not a Permutation (with 2 different letters and a Permutation, we would have 2 codes - but again, that's not allowed here).

GMAT assassins aren't born, they're made,
Rich

Excuse me, but the problem says "alphabetical order".
As I get it "alphabetical" means:
AB
BC
CD
DE
etc., where all letters come in an alphabetical sequence.

AB - is an "alphabetical order"
AC, AD, AE, etc - is not very "alphabetical". There are not so many alphabets with such an order of letters.

Very unclear problem imho.

Hi Antharas,

From what you mention in your post, I think that you might be confusing "alphabetical order" with "consecutive letters of the alphabet in order."

For example, with the letters A, B, C, D, E, there are a variety of pairs of letters that you could put in alphabetical order:
AB, AC, AD, AE, BC, BD, BE, CD, CE and DE

However, there are only 4 pairs that are consecutive letters in order:
AB, BC, CD and DE

Since the question is written in English, it's convenient to use the English alphabet as a reference, but you could use ANY alphabet (since there will be an 'alphabetical order' to it regardless of the country of origin).

GMAT questions are always carefully written, so you have to be mindful not to accidentally add any 'restrictions' to the information that prompt gives you.

GMAT assassins aren't born, they're made,
Rich

Hey,

I have a question regarding this. I decided not to use the regular formula and do it by trying. So my codes would be:

ZZ
YY
YZ
XX
XY
XZ
WW
WX
WY
WZ
Z
Y

I got 12 codes, all different, with only four letters. Can you please tell me what am I misinterpreting?

Hi madalenafnunes,

The prompt tells us that the potential codes can either be a single letter or two DISTINCT letters written in Alphabetical order. This means that the letters must be DIFFERENT (for example, XX is NOT an option) and 'order' (for example, XY is an option but YX is NOT). Your list includes 4 codes of duplicate-letters that are not allowed.

GMAT assassins aren't born, they're made,
Rich

Bunuel , VeritasKarishma , does the word pair have no significance here? i.e. can we even take a combination like ABC,BCD,CDE? Please advise.

No, you cannot take ABC as a code.

We are given that the code should be either a single letter so A or B or C etc or a pair of distinct letters written in alphabetical order so AB, AC, BC are acceptable codes but BA, ABC, AA are not.

So the question is that we should be able to make at least 12 total codes with the letters we pick.
If we pick 4 letters (A, B, C, D) , we can make 4 (single letter codes) + 4C2 (two letter codes) = 10 codes only.
Note that 4C2 gives you the number of codes with 2 letters. You do not need to arrange them because once you pick the letters, only one arrangement is possible since the letters must be placed alphabetically. So if I pick B and D, the only two letter code I get is BD.

If we pick 5 letters, we can make 5 + 5C2 = 15 codes.

Hence 5 letters are enough.

Hi Jitu20,

Yes - the word "pair" is significant because it defines the limitations on which codes are possible. The prompt mentions two options (separated by the word "or" in the sentence):

A code can consist of:
1) a single letter (such as A or B or C, etc.)
2) a pair (meaning 'two') of distinct (meaning different) letters written in alphabetical order (such as AB or AC or AD or BC, but NOT BA nor CA nor DB, etc.).

GMAT assassins aren't born, they're made,
Rich

If it was mentioned in the question that the alphabetically placed codes had to be consecutive, then the answer would be (D) 7 right? This is exactly what I assumed when solving first and got it wrong.

Hi LamboWalker,

The word "consecutive" does NOT appear in the original prompt. From what you mention in your post, I think that you might be confusing "alphabetical order" with "consecutive letters of the alphabet in order."

As an example, with the letters A, B, C, D, and E, there are a number of different pairs of letters that you could put in alphabetical order:
AB, AC, AD, AE, BC, BD, BE, CD, CE and DE.

In contrast, there are only 4 pairs that are consecutive letters in order (but this is NOT what the prompt described):
AB, BC, CD and DE

Since the question is written in English, it's convenient to use the English alphabet as a reference, but you could use ANY alphabet (since there will be an 'alphabetical order' to it regardless of the country of origin).

GMAT questions are always carefully written, so you have to be mindful not to accidentally add any 'restrictions' to the information that the prompt gives you.

GMAT assassins aren't born, they're made,
Rich

Kindly explain - why nc2 + n. I could not understand why you added the n.

I LOVE permutations and combinations. I studied combinatorics and graph theory in college. If this were a forum about those topics, I'd gladly catch you up. But this is a forum about the GMAT, so let's not waste your time. Twenty years ago, GMAC drastically reduced the weight that it applies to combinations and permutations. They are barely tested nowadays, and when they are, it's not this cumbersome. If you find an OFFICIAL combinations or permutations question from the past ten years that you have trouble with, tag me and I'll help you out. But only if there isn't a super-easy way to avoid the "real math" and still get the question right.

38% of people have missed this question on this site. THIRTY-EIGHT!!!! On a question that a ten-year-old could get right just by listing the options! Why are we making things more complicated than they need to be? Do we get bonus points for knowing the "real math?" No. Get the question right and move on.

A
B
C
D
AB
AC
AD
BC
BD
CD

That's not enough. Add E and we can tell it will be enough without even listing any more options. Done.

Seeking Help

I am curious why the order of the same letters cannot be reversed? For example AB and BA should considered as two pairs right? So it would be 2P4 instead of 2C4

Could anyone help me out?

The red section of the question stem requires the letter to be in alphabetical order. BA is not an option.

### Understanding the Problem
You need to find the smallest set of letters that can create unique codes for 12 participants. Codes can be a single letter or a pair of distinct letters in alphabetical order.

### Strategy
1. **Single Letters**: Each single letter can be a code by itself. So, if you have \(N\) letters, you get \(N\) codes right off the bat.
2. **Pairs of Distinct Letters**: If you have \(N\) letters, how many ways can you pair them up, ensuring they are distinct and in alphabetical order? This is a classic combination problem, where order doesn't matter, and you can't repeat letters. The formula for combinations is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items (letters, in this case), and \(k\) is the number of items to choose (2 for pairs).

### Goal
Find the smallest \(N\) such that the total number of codes (\(N + \text{combinations of N taken 2 at a time}\)) is at least 12.

### Calculation
1. **Single Letters**: \(N\) codes.
2. **Pairs of Letters**: \(\binom{N}{2}\) codes.

You need \(N + \binom{N}{2} \geq 12\).

Let's start with the smallest possibilities and work our way up:

- **For \(N = 4\)**: You get 4 single-letter codes. For pairs, \(\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6\) pairs. Total = \(4 + 6 = 10\) codes. That's not enough for 12 participants.
- **For \(N = 5\)**: You get 5 single-letter codes. For pairs, \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\) pairs. Total = \(5 + 10 = 15\) codes.

### Answer
With \(N = 5\), you get 15 unique codes, which is more than enough for 12 participants, making it the least number of letters needed. So, the answer is:

B. 5