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Will must choose a 3-character computer password, consisting of 1 lett
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04 Aug 2016, 10:12

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D

E

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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?

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 09:34

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5

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

Will must choose a 3-character computer password, consisting of 1 lett
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Updated on: 26 Jan 2020, 06:28

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abhimahna wrote:

GMATPrepNow 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

Can you please explain the 2nd scenario? Why can't we take 10c1 * 9c1 for selecting the two digits? I think I am missing something here. Please elaborate.

You are treating stage 2 as though the order in which we select the 2 digits matters, when the order does not matter (we arrange the letters in stage 3). So, in your solution, selecting digit 5 first and then selecting 7 second is treated as an outcome that's DIFFERENT from selecting digit 7 first and then selecting 5 second , when both of these outcomes are IDENTICAL.

Re: Will must choose a 3-character computer password, consisting of 1 lett
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26 Jul 2017, 15:35

1

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.

Re: Will must choose a 3-character computer password, consisting of 1 lett
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25 Aug 2018, 05:31

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jackjones wrote:

Sorry, I don't understand why the two digits to be used is 10C2, if the first digit can be selected in 10 ways and the second d one in 10 ways then should it not be 10x10? what am I missing?

There are various ways to answer this question. In my solution, I first chose the 3 characters (disregarding the order) and THEN I arranged the 3 characters I had chosen.

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Nov 2019, 11:45

1

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:

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....

Re: Will must choose a 3-character computer password, consisting of 1 lett
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26 Jan 2020, 08:00

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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.
_________________

Will must choose a 3-character computer password, consisting of 1 lett
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Updated on: 05 Aug 2016, 06:33

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

26C1*10C1*9C1 = 2340.

=> 2340 * 3 ! = 14,040.

Option E is correct answer...but OA is D.

Abdur..can you give us official solution...will correct if I missed anything..

Originally posted by msk0657 on 05 Aug 2016, 06:10.
Last edited by msk0657 on 05 Aug 2016, 06:33, edited 1 time in total.

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 06:24

msk0657 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

26C1*10C1*9C1 = 2340.

=> 2340 * 3 = 7020.

Option D is correct answer.

Why have you not multiplied 2340 with 3!. It is a three character password, so order of the characters does matter. I am not sure but I am getting the OA as E.
_________________

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 06:29

abhimahna wrote:

msk0657 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

26C1*10C1*9C1 = 2340.

=> 2340 * 3 = 7020.

Option D is correct answer.

Why have you not multiplied 2340 with 3!. It is a three character password, so order of the characters does matter. I am not sure but I am getting the OA as E.

Yes... you are correct I missed 3! ...then it'll be E... I'll update..

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 08:37

msk0657 wrote:

abhimahna wrote:

msk0657 wrote:

26C1*10C1*9C1 = 2340.

=> 2340 * 3 = 7020.

Option D is correct answer.

Why have you not multiplied 2340 with 3!. It is a three character password, so order of the characters does matter. I am not sure but I am getting the OA as E.

Yes... you are correct I missed 3! ...then it'll be E... I'll update..

Thanks , but my concern is why the overall answer marked as D here?

Can someone please explain where I am going wrong?
_________________

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 09:49

1

GMATPrepNow 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

Can you please explain the 2nd scenario? Why can't we take 10c1 * 9c1 for selecting the two digits? I think I am missing something here. Please elaborate.
_________________

Re: Will must choose a 3-character computer password, consisting of 1 lett
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05 Aug 2016, 10:23

GMATPrepNow wrote:

abhimahna wrote:

GMATPrepNow wrote:

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

Can you please explain the 2nd scenario? Why can't we take 10c1 * 9c1 for selecting the two digits? I think I am missing something here. Please elaborate.

You are treating stage 2 as though the order in which we select the 2 digits matters, when the order does not matter (we arrange the letters in stage 3). So, in your solution, selecting digit 5 first and then selecting 7 second is treated as an outcome that's DIFFERENT from selecting digit 7 first and then selecting 5 second , when both of these outcomes are IDENTICAL.

This video might help:

Cheers, Brent

Sounds Cool, Thanks Brent.

May be I am too tired now that I didn't recognize such a simple thing. But your answer made me realize my mistake. Thanks
_________________

Re: Will must choose a 3-character computer password, consisting of 1 lett
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24 Jul 2017, 17:35

GMATPrepNow 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

Re: Will must choose a 3-character computer password, consisting of 1 lett
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24 Jul 2017, 18:13

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septwibowo wrote:

GMATPrepNow 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

Re: Will must choose a 3-character computer password, consisting of 1 lett
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27 Jul 2017, 17:13

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

Order of the Digits don't matter?

or is because we took 9x10, we are already accounting for the order.

Re: Will must choose a 3-character computer password, consisting of 1 lett
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25 Aug 2018, 03:49

GMATPrepNow 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

Sorry, I don't understand why the two digits to be used is 10C2, if the first digit can be selected in 10 ways and the second d one in 10 ways then should it not be 10x10? what am I missing?

Will must choose a 3-character computer password, consisting of 1 lett
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26 Jan 2020, 04:37

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:

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....

Re: Will must choose a 3-character computer password, consisting of 1 lett
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26 Jan 2020, 06:32

Top Contributor

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:

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....

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!