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

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Manager
Joined: 27 Mar 2017
Posts: 120
Re: Will must choose a 3-character computer password, consisting of 1 lett  [#permalink]

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26 Jan 2020, 07:18
GMATPrepNow 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

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

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

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
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Joined: 11 Sep 2015
Posts: 4547
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Re: Will must choose a 3-character computer password, consisting of 1 lett  [#permalink]

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26 Jan 2020, 08:00
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Top Contributor

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]

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26 Jan 2020, 17:19
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
Manager
Joined: 27 Mar 2017
Posts: 120
Re: Will must choose a 3-character computer password, consisting of 1 lett  [#permalink]

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28 Jan 2020, 22:51
GMATPrepNow wrote:

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]

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28 Jan 2020, 23:01
\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)

$$= 26C1 * 10C2 * 3!$$

$$= 26 * 45 * 6$$

$$= 7020$$

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Will must choose a 3-character computer password, consisting of 1 lett   [#permalink] 28 Jan 2020, 23:01

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