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Will must choose a 3-character computer password, consisting of 1 lett

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altairahmad
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  26 Jan 2020, 08:18
GMATPrepNow wrote:
altairahmad wrote:
EMPOWERgmatRichC wrote:
Hi All,

We're told that Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order. We're asked for the total number of different passwords that Will can choose. Depending on how comfortable you are with the 'math' involved, you can perform the calculations in a couple of different ways. Here's how you can break the prompt down into smaller pieces (which you might find easier than trying to do one gigantic calculation).

Based on the the 'restrictions' in the prompt, we can use 1 letter of the English alphabet and 2 DISTINCT digits IN ANY ORDER to make a code. Thus, the code could be one of 3 options:

(Letter)(Digit)(Different Digit)
(Digit)(Letter)(Different Digit)
(Digit)(Different Digit)(Letter)

The first option = (Letter)(Digit)(Different Digit) = (26)(10)(9) = 2340
The second option = (Digit)(Letter)(Different Digit) = (10)(26)(9) = 2340
The third option = (Digit)(Different Digit)(Letter) = (10)(9)(26) = 2340

You might recognize that each calculation involves the product of the same 3 numbers, so you don't have to do that calculation each time - just do it once and then multiply that result by 3....

Total options = 3(2340) = 7020 options

Final Answer:
Show Spoiler
D


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


and GMATPrepNow Bunuel please advise regarding below confusion

Why didn't we multiply 2340 with 3! ? Afterall, A31 is different from A13. Isn't it ? In LDD both Ds are distinct and its basically LD1D2 situation ?


Good question.
In my approach (at https://gmatclub.com/forum/will-must-ch ... l#p1719321), I first chose two digits and one letter, and then rearrange this collection of three characters in 3! ways.
If we use this approach, then we must recognize that the order in which we select our two digits does not matter, which means we can select two distinct digits in 10C2 ways (45 ways).

In Rich's approach shown above, we are not following the same approach that I did.
Instead we are treating each case differently.
As such we don't need to multiply by 3!

Cheers,
Brent


Thanks for the reply.

My question is precisely about the last part of your post. How are there 3 situations only ?

LDD
DDL
DLD

Taking the last one DLD e.g implies 1A3 is same as 3A1 whereas it is not the case. So shouldnt we multiply with 3! Instead of just 3.

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BrentGMATPrepNow
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  26 Jan 2020, 09:00
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altairahmad wrote:
Thanks for the reply.

My question is precisely about the last part of your post. How are there 3 situations only ?

LDD
DDL
DLD

Taking the last one DLD e.g implies 1A3 is same as 3A1 whereas it is not the case. So shouldnt we multiply with 3! Instead of just 3.

Posted from my mobile device


Let's examine the DLD configuration.
In Rich's solution, there are 10 ways to select the first character (0,1,2,...9), there are 26 ways to select the second character (A,B,C,D,...Z)) and there are 9 ways to select the third character.
At this point we have accounted for ALL 2340 possible outcomes in a DLD configuration.
So there's no need two multiply anything by 3!

NOTE: The above approach does not imply that 1A3 is same as 3A1

Likewise, there are 2340 possible outcomes in a LDD configuration, and 2340 possible outcomes in a DDL configuration.

So once we consider these three possible configurations, we are done.

Conversely, in my solution, we first select two digits and one letter, and THEN we arrange them in 3! different ways.
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  26 Jan 2020, 18:19
There are 3! ways to combine three items.
Since two are digits, that means we have repeats.
So we have 3!/2! = 3 variations

total spread is 26*10*9 = 2340
2340*3 = 7,020
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  28 Jan 2020, 23:51
GMATPrepNow wrote:
altairahmad wrote:
Thanks for the reply.

My question is precisely about the last part of your post. How are there 3 situations only ?

LDD
DDL
DLD

Taking the last one DLD e.g implies 1A3 is same as 3A1 whereas it is not the case. So shouldnt we multiply with 3! Instead of just 3.

Posted from my mobile device


Let's examine the DLD configuration.
In Rich's solution, there are 10 ways to select the first character (0,1,2,...9), there are 26 ways to select the second character (A,B,C,D,...Z)) and there are 9 ways to select the third character.
At this point we have accounted for ALL 2340 possible outcomes in a DLD configuration.
So there's no need two multiply anything by 3!

NOTE: The above approach does not imply that 1A3 is same as 3A1

Likewise, there are 2340 possible outcomes in a LDD configuration, and 2340 possible outcomes in a DDL configuration.

So once we consider these three possible configurations, we are done.

Conversely, in my solution, we first select two digits and one letter, and THEN we arrange them in 3! different ways.


I have understood the flaw in my reasoning. Thanks a lot for explaining.
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Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  29 Jan 2020, 00:01
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\sqrt{}
AbdurRakib wrote:
Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order.From how many different passwords can Will choose?

A. 390
B. 2,340
C. 4,680
D. 7,020
E. 14,040



We need to fix 3 characters - 1 letter from 26 available letters and 2 digits from 10 available digits. Also, the characters can be arranged in any order, i.e. X10 is different from 0X1 and so on.

We can read the question as (to make it easy to understand):

Select 1 letter AND Select 2 digits AND Arrange these 3 distinct characters

Thus, number of ways (remember: AND means you MULTIPLY; OR means you ADD)

\(= 26C1 * 10C2 * 3!\)

\(= 26 * 45 * 6\)

\(= 7020\)

Answer D
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  04 Sep 2020, 10:04
ScottTargetTestPrep wrote:
AbdurRakib wrote:
Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order.From how many different passwords can Will choose?

A. 390
B. 2,340
C. 4,680
D. 7,020
E. 14,040


Since there are 26 letters and 10 digits, the number of 3-character passwords that can be created is 26 x 10 x 9 = 2,340, if the password is in the form of LDD where L means letter and D means digit. However, the password can be also in the form of DLD and DDL, each of which also can be created in 2,340 ways. Thus, the total number of passwords is 2,340 x 3 = 7,020.

Answer: D


ScottTargetTestPrep Hello Scott I was wondering whether you could clarify a step on your approach, I was wondeering why did you multiply with 3 and not with 6 the password is in the form LD1D2 , and we care about order , what am I missing here?
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  08 Sep 2020, 06:05
BrentGMATPrepNow wrote:
AbdurRakib wrote:
Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order.From how many different passwords can Will choose?

A. 390
B. 2,340
C. 4,680
D. 7,020
E. 14,040


Take the task of creating a password and break it into stages.

Stage 1: Select the one letter to be used in the code
There are 26 letters from which to choose, so we can complete this stage in 26 ways.

Stage 2: Select the two digits to be used in the code
Since the order in which we select the two digits does not matter, we can use combinations.
We can select 2 digits from 10 women in 10C2 ways (45 ways)
So, we can complete stage 2 in 45 ways

NOTE: We now have the 3 characters to be used in the code. At this point, we need to arrange those 3 characters.

Stage 3: Arrange the 3 selected characters.
RULE: We can arrange n unique objects in n! ways.
So, we can arrange the 3 characters in 3! ways (6 ways)
So we can complete stage 3 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a password ) in (26)(45)(6) ways
(26)(45)(6) = 7020

Answer:
Show Spoiler
D


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Hello Brent,
Blast from the past here, please do help. How does one safely go about assuming that order doesn't matter in a certain question?
The question here says "in any order", I assumed that to mean that the position of the letter and the digits is not fixed, i.e. there are no constraints on where exactly the letters or digits must appear.
Considering that in the real world, A91 is certainly a different password than 9A1, how do we assume that 10C2 is right and not 10C1*9C1?
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BrentGMATPrepNow
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  08 Sep 2020, 07:34
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ShreyasJavahar wrote:
Hello Brent,
Blast from the past here, please do help. How does one safely go about assuming that order doesn't matter in a certain question?
The question here says "in any order", I assumed that to mean that the position of the letter and the digits is not fixed, i.e. there are no constraints on where exactly the letters or digits must appear.
Considering that in the real world, A91 is certainly a different password than 9A1, how do we assume that 10C2 is right and not 10C1*9C1?


Great question!
Perhaps a few extra words should be added to the question to make it less ambiguous.
For example, we could add that A91 and 9A1 are both permissible passwords, AND they're considered different outcomes.
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ScottTargetTestPrep
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  08 Sep 2020, 19:22
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UNSTOPPABLE12 wrote:
ScottTargetTestPrep wrote:
AbdurRakib wrote:
Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order.From how many different passwords can Will choose?

A. 390
B. 2,340
C. 4,680
D. 7,020
E. 14,040


Since there are 26 letters and 10 digits, the number of 3-character passwords that can be created is 26 x 10 x 9 = 2,340, if the password is in the form of LDD where L means letter and D means digit. However, the password can be also in the form of DLD and DDL, each of which also can be created in 2,340 ways. Thus, the total number of passwords is 2,340 x 3 = 7,020.

Answer: D


ScottTargetTestPrep Hello Scott I was wondering whether you could clarify a step on your approach, I was wondeering why did you multiply with 3 and not with 6 the password is in the form LD1D2 , and we care about order , what am I missing here?


The term LDD means that the password is in the form of a letter, followed by a digit, followed by another digit. Similarly for DLD and DDL.
It is impossible to distinguish "LD1D2" from "LD2D1"; in either case, the possible values for D1 and D2 are the digits between 0 and 9. If there was a way to distinguish one digit in the password from the other digit, then we would have to consider separate cases of LD1D2 and LD2D1. For instance, if the question told us that the password should contain a letter, one digit from the digits 0 to 9 and one digit from the Thai alphabet ๐ ๑ ๒ ๓ ๔ ๕ ๖ ๗ ๘ ๙, then we would have to consider LD1D2 and LD2D1. Notice that in this modified question, the password "A One Nine" can appear in two distinct ways, as A1๙ or as A๑9. But since both digits we use for the password in the original question are 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9, we should only consider the cases LDD, DLD and DDL.
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BrentGMATPrepNow
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Re: Will must choose a 3-character computer password, consisting of 1 lett [#permalink] New post  04 Dec 2020, 07:12
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AbdurRakib wrote:
Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order. From how many different passwords can Will choose?

A. 390
B. 2,340
C. 4,680
D. 7,020
E. 14,040


Here's another approach.

There are three possible cases that satisfy the given information.
case i: The password is in the form letter-digit-digit
case ii: The password is in the form digit-letter-digit
case iii: The password is in the form digit-digit-letter

case i: The password is in the form letter-digit-digit
There are 26 ways to select the only letter
There are 10 way to select the first digit
There are 9 ways to select the other digit (it can be any digit except the digit we chose for second position)
So the total number of passwords in the form letter-digit-digit = (26)(10)(9) = 2340

case ii: The password is in the form digit-letter-digit
There are 10 ways to select the first digit (0,1,2,3,4,5,6,7,8 or 9)
There are 26 ways to select the only letter
There are 9 way to select the other digit (it can be any digit except the digit we chose for first position)
So, the total number of passwords in the form digit-letter-digit = (10)(26)(9) = 2340

As you might guess, the third case can also be accomplished in 2340 ways

Answer: (3)(2340) = 7020

Answer: D

Cheers,
Brent
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