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Difficulty:
65%
(hard)
Question Stats:
60% (02:10) correct
40%
(02:10)
wrong
based on 925
sessions
History
Date
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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!
_________________
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!
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Total number of four letter passwords = 6*6*6*6= 1296 ------(1)
Total number of passwords in which no letter repeats = 6C4*4!= 15*24=360------(2)
therefore required value = (1)-(2)= 1296-360=936
Total number of passwords in which no letter repeats = 6C4*4!= 15*24=360------(2)
therefore required value = (1)-(2)= 1296-360=936
Bunuel
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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General Discussion
4 places 6 alpha, total number of ways in which a code can be formed without any restrictions = 6x6x6x6 = 1296
Now question says the code can be formed with at least 1 repetition so here shortcut = total - code with no repetition.
1296 - (code with no rep = 6x5x4x3 =) 360 = 936
Now question says the code can be formed with at least 1 repetition so here shortcut = total - code with no repetition.
1296 - (code with no rep = 6x5x4x3 =) 360 = 936
Alex75PAris
Joined: 16 Mar 2016
Last visit: 08 Mar 2017
Posts: 98
Location: France
Schools: Marshall '19 Darden '19 HKUST '19 Anderson '19
GMAT 1: 660 Q47 V33
GPA: 3.25
Total number of possibilities - Number of possibilities without repetition
6*6*6*6 - 6*5*4*3 = 1296 - 360 = 936
6*6*6*6 - 6*5*4*3 = 1296 - 360 = 936
Posts: 27
Bunuel
What is the meaning of "and/or F"
Posted from my mobile device
Posts: 1,070
Updated on: 05 Aug 2022, 16:27
Originally posted by ThatDudeKnows on 05 Aug 2022, 16:24.
Last edited by ThatDudeKnows on 05 Aug 2022, 16:27, edited 1 time in total.
Last edited by ThatDudeKnows on 05 Aug 2022, 16:27, edited 1 time in total.
SolankiDas
SolankiDas
If it used only "and," some people might interpret that to mean that all six letters must be used. But there are only four slots.
If it used only "or," some people might interpret that to mean that only one letter is used and that it is repeated four times. But that's not possible given the other constraint.
By using "and/or," they're just trying to alleviate the ambiguity of each of those words individually.
IMO, it's a bit of a sloppy way to clear the ambiguity and I wouldn't expect to see GMAC use the same wording, so don't sweat it!
Posts: 1,070
Bunuel
I was with the official answer right up until the part in red. Here's how I'd have gone about this question starting from the beginning.
6^4 is going to end in a 6, right?
6*5*4*3 is going to end in 0, right?
We have something that ends in a 6 minus something that ends in a 0, so we need an answer that ends in a 6. Eliminate A, B, and C.
At this point, we need to eliminate one more. 6^4 is 1296. And then we are going to subtract something. Eliminate E.
Answer choice D.
Super quick 30 secs method:
Digit is repeated atleast once = All possible cases - no digit is repeated
6x6x6x6 - 6x5x4x3
6 (6^3 - 5x4x3) = 6(216-60) = 6x156 = 936 D
Digit is repeated atleast once = All possible cases - no digit is repeated
6x6x6x6 - 6x5x4x3
6 (6^3 - 5x4x3) = 6(216-60) = 6x156 = 936 D
Posts: 5
Bunuel
This is definitely a question that shows my fundamental understanding isn't yet solid, but how come you calculate 6^4 for all of the password options. What principle is that assumption based upon?
Total number of four letter passwords = 6^4 i.e 6*6*6*6= 1296
Total number of passwords in which no letter repeats = 6C4*4!= 15*24
Here is a faster calcluation tip using units digit
Notice the units digit of 6^4 =6
The units digit of 15 x 24 = 0
Hence on subtraction the solution will have a units digit of 6
Only answer choices D and E have the units digit of 6 but E cannot be the answer cause 6^4 = 1296 | Memorize this value it comes in handy
Hence mark D hence no need to even multiply.
Total number of passwords in which no letter repeats = 6C4*4!= 15*24
Here is a faster calcluation tip using units digit
Notice the units digit of 6^4 =6
The units digit of 15 x 24 = 0
Hence on subtraction the solution will have a units digit of 6
Only answer choices D and E have the units digit of 6 but E cannot be the answer cause 6^4 = 1296 | Memorize this value it comes in handy
Hence mark D hence no need to even multiply.
6^4-(5x4x3x2) = 936
Bunuel
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