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

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?

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.

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

Dear GMATPrepNow,

Can we solve this question using this approach?

1. We define 3 stages : one slot must be filled with one from 26 alphabets and 2 slot must be filled with two numbers.

2. From this 3 stages, I break again into three possibility :

- First : Alphabet in the first digit
Number of possible password : 26 X 10 X 9 = 2.340

- Second : Alphabet in the middle digit
Number of possible password : 10 X 26 X 9 = 2.340

- Third : Alphabet in the last digit
Number of possible password : 10 X 9 X 26 = 2.340

3. We sum up total possibilites, and we get = 7.020.


Please correct my approach.

Thanks.

That's a perfectly valid approach - nice work!

Cheers,
Brent

No. of ways of choosing an alphabet = 26c1 = 26.

No. of ways of choosing 2 digits from 0-9 = 10c2 = 45.

Since the characters in the password can be in any order, the alphabet can be placed X-0-0, or 0-X-0, or 0-0-X = in 3 ways.

Similarly, the numbers can be placed in 2! ways.

Total no. of passwords = 2! * 3 * 26 * 45 = 7020.

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.

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.

Does that help?

Cheers,
Brent

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:

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

possible arrangements ; 26 for alphabet and 10*9 for digits
total ways ; 3
26*10*9*3 ; 7020
IMO D