Dec 18, 2018

There's a quote attributed to the famous Renaissance artist Michaelangelo, supposedly in response to a fan who asked Michaelangelo how he was able to sculpt such a masterpiece in "David" out of a formless block of marble, that goes as follows:

**"[Sculpting David] is easy. You just chip away the stone that doesn't look like David."**

Whether that's an actual Michaelangelo quote or not, I like the sentiment -- and it applies perfectly to certain question types that you make encounter on the GMAT or GRE.

You may remember the complement rule of probability that states,

**probability (something) = 1 - probability (not something)**

In other words, probabilities must sum to 100%, so if the probability of something happening is 60%, for example, then the probability of that thing *not* happening is 40% (1 - .60 = .40). Pretty straightforward.

Now as I explain in the probability lessons of my GMAT and GRE courses, this formula is particularly helpful on hard probability questions that ask you to find the chance that something will happen "at least" or "at most" a certain number of times. The idea is that instead of calculating a bunch of independent probabilities that satisfy what the question is asking for and then adding them together, **it's often faster and easier to think about what you don't want (i.e. what isn't considered a favorable outcome), and then subtract it from the total (often 1, or 100%).**

Sort of like Michaelangelo chipping away what he *didn't* want, leaving only what he *did* want -- David!

Now, this way of thinking isn't only helpful on hard probability questions. Indeed, any time you see "at least" questions, even on combinatorics ("how many ways...?") problems, you can use the complement rule.

Consider the following example:

*Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?A. 2,145B. 6,435C. 12,100D. 12,870E. 25,740*

At first glance, this question looks like a doozy. And it is pretty challenging. But let's look at how much easier it becomes when we apply the logic discussed above.

When we think about this question in terms of the complement rule, we recognize that:

**3-letter passwords with at least one symmetrical letter = (All the combinations of 3-letter passwords) - (3-letter passwords with not at least one symmetrical letter)**

In other words, rather than worrying about directly figuring out which combinations constitute "at least one symmetrical letter," we can instead figure out how many passwords would not contain a symmetrical letter, and then subtract that from the total.

Okay, so that's a start. But how do we interpret the "not at least one symmetrical letter" part? Hmmm.... Ah! That's the same as a situation where we have only asymmetrical letters, right? In other words, that's the only thing we *don't* want. Everything else counts for what we're looking for. But if we have a password with only asymmetrical letters, then the password doesn't have "at least one" symmetrical letter, so it's to be excluded.

Now let's rewrite our formula to get crystal clear on what we're trying to calculate, using this new wording:

**3-letter passwords with at least one symmetrical letter = (All the combinations of 3-letter passwords) - (3-letter passwords with all asymmetrical letters)**

That we can handle, I think.

Let's calculate each part of the formula in turn:

*All the combinations of 3-letter passwords*=**26 * 25 * 24 = 15,600.**(There are 26 letters in the English alphabet, and any could be the first letter which is why you start with 26; then any of the remaining 25 could be second; then any of the remaining 24 could be third.)**Password combinations with all asymmetrical letters****= 15 * 14 * 13 = 2,730**. (There are 15 asymmetrical letters, which is why you start with 15, and then you continue as explained above.)

Therefore applying the complement rule, **the number of 3-letter passwords with at least one symmetrical letter =** **15,600 - 2,730 = 12,870**.

Answer choice D.

Now, we could have tackled this question the direct way. But as we'll see, it's quite a bit more involved, takes a lot more time, and has more places where you could possibly make an error and therefore get a wrong answer.

Nevertheless, let's give it a try. The direct approach requires us to think about what we *do* want. In other words, what counts as a scenario where we have a password with "at least one" symmetrical letter? Well, there are three such cases:

- All 3 letters are symmetrical
- 2 of the letters are symmetrical, the other 1 is asymmetrical
- 1 of the letters is symmetrical, the other 2 are asymmetrical

Each of those password combinations has "at least 1" symmetrical letter. As we noted previously, the only scenario that doesn't count would be a password with all 3 asymmetrical letters.

So now we need to figure out how many permutations there are of each of those three scenarios, and then add them together. Doing so requires a bit more advanced understanding of combinatorics, so if you don't understand what I'm about to explain, check out our GMAT quant course or GRE quant course. I cover advanced combinatorics and probability concepts in detail there.

But here's how the math works:

- All 3 letters are symmetrical = 11P3 = 990
- 2 of the letters are symmetrical, the other 1 is asymmetrical = 11P2 * 15 = 110 * 15, but there are 3 different ways to have the 11P2, so it's (110 * 3) * 15 = 4,950
- 1of the letters is symmetrical, the other 2 are asymmetrical = 11 * 15P2 = 11 * 210, but there are 3 different ways to have the 15P2, so it's 11 * (210 * 3) = 6,930

Finally you add them all together and you get 990 + 4,950 + 6,930 = 12,870. Same answer as above, but a lot more steps.

Students are often inclined to dive head-first into quant problems on the GMAT and GRE and try to solve them in a traditional, straightforward way. But that's not always ideal. As you've seen, getting rid of what you *don't* want can sometimes leave exactly what you *do* want. That's a principle that extends beyond "at least" questions, by the way. For example, this same mindset is helpful on shaded region questions. It's also helpful when eliminating eye-catch wrong answers on both quant and verbal. But those are topics for another article. For now, hopefully this mindset shift will enable you to get more of these challenging quantitative question types correct, which will produce your own masterpiece on test day.

*What did you learn? What questions do you still have? Please leave a comment below!*

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