GMAT Math

All of the stocks on the over-the-counter market are designated by either a 4-letter or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes?

a. 2(26^5)
b. 26(26^4)
c. 27(26^4)
d. 26(26^5)
e. 27(26^5)

Hey Naveen,

Great question here - thanks for sharing all of these today! I'll let the community dig into most of these but I noticed the setup on this one and think it provides a great teaching platform.

Here, the answer choices are a huge part of the question...one great thing about combinatorics and probability problems is that the answers often come in an "unfinished" form, because the numbers themselves are huge. When that's the case, the answer choices themselves provide excellent blueprints as to the kind of math that you need to do, because they're given with an operation involved...they're telling you what the steps are that you'd need to take to get there.

Here, hopefully you'll get pretty quickly to noting that there will be 26^4 4-letter codes and 26^5 5-letter codes. But what do you do with that?

26^4 and 26^5

The answer choices are in factored-addition form, so that should give you a clue...26^9 would be a great trap answer choice for those who want to immediately multiply those numbers together, but in actuality there are 26^4 4-letter codes and 26^5 5-letter codes, and one doesn't depend on the other...you'd never use one of the 4-codes and one of the 5-codes simultaneously, so logically we have to add those together. The total number of available codes is going to be the sum of one type of code and the other type of code: 26^4 + 26^5.

26^4 + 26^5

Now, with that process in mind, we have to make our answer choice look like theirs. And theirs has a term in parentheses. The main way to do that with an addition problem is to factor out a common term. both of these have a 26^4 in them, so you can factor that to:

(26^4)(1 + 26)

And if we sum the simple addition in the second parentheses, we'll have:

(26^4)(27)

And that matches choice C.


Looking back at this question, it gives you some clues. Not only does it have the parentheses in there to demonstrate that you'll probably need to factor out some addition, but two choices have the presence of the number 27 in them...and that's not a "natural" number in front of you. If you quickly jumped to B or D because of that 26, you might want to go back and ask yourself "how would someone even get 27 out of this?". Because often if a large-number calculation (exponents, factorials, etc.) seems to lend itself to an "obvious" answer that you get in 20 seconds or less, you may be missing a piece of embedded difficulty that doesn't catch the eye immediately. The answer choices can clue you in to that - each answer choice, particularly if it's in a calculation form and not just a number, is crafted so that someone would pick it. So if multiple answer choices elude you as to how anyone in their right mind might come to that number, it may be a decent indication that your 20-second answer is a bit too simple and may require a little extra thought.

One well-kept secret of most GMAT math problems is that the types of answer choices necessarily have to give you clues. Learn to read those clues and you have that much more information at your disposal...

Thanks naveenhv, for the problem, and thanks to Brian@VeritasPrep, for explaining to us the way to solve it

Hey Brian,

Thanks for explaining the answer.

I have posted all the question i got wrong on my GMAT Prep test on Sunday. So there are quite a few from my side today.

@kingluis, you are most welcome. Hopefully, this question helps folks to learn something new.

I need clarification on this question. If you need to solve four-letter groups out of the total of 26, shouldn't you use the combination formula? I don't get why it's 26^4. (Obviously, math is not my strong suit.) An elaboration would be much appreciated, thanks!

Hey glasshalffull - great question (and I love the optimism in your username!).

If I can get on my teacher's soapbox for a second, this is where I hate the way that most people study. The GMAT is decidedly not a test of "do you know the formulas" - for the rest of your life you'll have Google on your smartphone and a team of Ivy League interns to know all that stuff for you. So it's not really a formula-based test. Thinking beats remembering just about every time on the GMAT, so if you're ever thinking "I should just plug this into 'the formula'", ask yourself why that formula applies and whether there's any special circumstance to it.

The permutations formula is N!/[(N-K)!], and that applies when you need to select K unique items from a set of N unique items, and and the order matters (arrangements of items, typically).

The combinations formula is N!/[K!(N-K)!] and applies when you need to select K unique items from a set of N unique items and the order does not matter.

Here, the order does matter, right? AAAB is a totally different code from AABA. So because the order of the items matters, we wouldn't use the combinations formula.

We also wouldn't use the permutations formula. Why not? Because we're not using unique items - as we saw with AAAB vs. AABA, we can use the same letter multiple times. What we're really calculating - formula or no formula - is how many options we have for each slot. And for this one, we have:

For a 4-letter code:

The first slot: 26 available options
The second slot: 26 avialable options
The third slot: 26 available options
The fourth slot: 26 available options

And we multiply those together, because for each of the 26 first-letter options, we can branch out into 26 unique second options, then for each of those there are 26 third options, etc.

So the number of 4-letter codes is 26*26*26*26.

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Now here's how they'd ask it as a permutations problem that uses that formula:

How many 4-letter codes can be issued for stocks if each letter can only be used once in any given code?

Here, we'd have:

First letter: All 26 available options
Second letter: Any of the 25 that we didn't use for #1
Third letter: Any of the 24 that remain after the first 2
Fourth letter: Any of the 23 remaining letters after the first 3

or 26*25*24*23

We'd get that, too, by plugging N (26 unique items) and K (the number that we're going to use in our code) into that formula:

N! / [(N-K)!]
26! / [(26-4)!]
26! / 22!

Remember, 26! = 26*25*24*23*(22!), so the 22! in both numerator and denominator would factor to just leave 26*25*24*23.

Honestly, I'd argue that most combinatorics problems are easier to solve by thinking through the number of options than by just using the formula, and even those that would be better off via a formula tend to require some kind of thinking beyond just the formula, so you'll want to know some of the logic behind it anyway to be able to do that thinking.

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Alright, one more thing, and please don't take this as blatant salesmanship but if you're looking for a good book to click with this kind of thinking in a step-by-step way that takes you from basic combinatorics to complicated applications, my good friend and colleague, Chris, wrote most of the Veritas Prep Combinatorics and Probability book, and made the subject much, much clearer for me. https://www.amazon.com/Combinatorics-Pro ... _lmf_tit_5

Thanks, Brian! That was very helpful, especially as I'm about to venture into practicing mode. I'll make sure to maintain that mindset as I study.