Bunuel wrote:
In how many ways can a four-letter password be chosen, using the letters A, B, C, D, E, and/or F, such that at least one letter is repeated within the password?
A. 720
B. 864
C. 900
D. 936
E. 1,296
You have 1 minute (yes, only one minute!) to do this problem. Further, don’t do any “long-hand” math (multiplication, etc.)—shortcuts exist for every single calculation. Find them!
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:Why did we push you to do this problem so quickly? It wasn’t because we want you to work fast just for the sake of working fast. (And please do NOT take the message away that you need to jump in and force yourself to plow ahead just for the sake of working fast.)
Rather, we want you to learn that there are often “elegant” solutions to hard math problems—solutions that take far less mental energy (and therefore less time) if you have learned how to use them. If you seek out these approaches during study, your task will be much easier when you take the real test.
It’s possible to calculate this result directly, but not in a minute (unless you’re a human calculator). You would need to enumerate not only all the different ways in which the password could be formed (one repeated letter, two different repeated letters, 3 instances of one letter, etc.), but also all the different orders of letters!
So the first lesson is this: when faced with a tedious solution approach, slow down (yes, pause for a few seconds of your precious minute!) and take a look at the wording of the problem again. Is there a better way to set this up?
The “at least” language signals that you can find the total number of ways to make any password, and then subtract out what you don’t want: the number of passwords in which letters are not repeated, a much easier calculation.
The total number of passwords altogether, with or without repeated letters, is
6*6*6*6 = 6^4
The number of passwords with no repeated letter is 6*5*4*3.
Now, think about how to save yourself some time in the calculation stage. The desired number of passwords is:
= 6^4 - 6*5*4*3 pull out a 6
= 6(6^3 - 5*4*3) at this level, you should have 6^3 memorized!
= 6(216-60) subtract the ten’s digit (21-6 = 15), then the unit’s digit to get 156
= 6(156) multiply 6 by 150 = 900, then add 6*6 = 36
= 936
But wait! You could make these calculations even easier. (You might not spot this the first time you try the problem, but you might pick this up while you’re analyzing the problem after the fact. Then you’ll know to keep your eyes open for these opportunities in future.)
Here’s how:
= 6^4 - 6*5*4*3
Hmm… 6^3= 216, so 6^4= roughly 200*6 = roughly 1,200. Anything else? The number 6 raised to any power has to end in 6, so that 1200 estimation is really 12x6 (where x represents the one digit we know nothing about).
= ~12x6 - 6*5*4*3
Okay, let’s do some of the math and estimate when it gets tough. 6*5 = 30*4 = 120*3 = 360. That was easier than it looked at first glance.
= ~12x6 - 360
= something in the 800 or 900 range, ending in 6
The only 800- or 900- range answer that ends in 6 is 936.
The correct answer is D.
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