Each student at a certain university is given a four-charact

Question

Each student at a certain university is given a four-character identification code, the rest two characters of which are digits between 0 and 9, inclusive, and the last two characters of which are selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?

A . 135,200
B. 67,600
C. 64,000
D. 60,840
E. 58,500

A. 135,200
B. 67,600
C. 64,000
D. 60,840
E. 58,500

Bunuel · Accepted Answer

karishmatandon · Answer

Four character identification code
_ _ _ _ 
First two parts for the code, are digits between 0-9, therefore, 10 options for the first part of the code,
and as characters may be repeated, 10 options for the second part as well
Therefore, we have 10 X 10 possibilities for the first and second part of the code

Last two parts of the code, are characters selected from the 26 letters of the alphabet, therefore, 26 options for the third part of the code,
and as characters may be repeated, 26 options for the fourth part as well
Therefore, we have 26 X 26 possibilities for the third and fourth part of the code

so, in all total no. of different identification codes generated following these rules
= 10 X 10 X 26 X 26 = 67600

Answer B

EgmatQuantExpert · Answer

Each student at a certain university is given a four-character identification code, the rest two characters of which are digits between 0 and 9, inclusive, and the last two characters of which are selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?

A. 135,200
B. 67,600
C. 64,000
D. 60,840
E. 58,500

The question here is asking us to fill four places with given set of letters and digits. We are given a constraint that first two places can only be filled with digits and the last two places can only be filled with letters. We are also told that the characters can be repeated. Let's see the number of ways in which each place can be filled.

1st Place: The 1st place of the code needs to be filled with digits only. The total number of digits which we have is 10 ( from 0 to 9 both inclusive). So, there are 10 ways in which we can fill the first place.

2nd Place: The 2nd place also needs to be filled with digits only. Since we are given that digits can be repeated, we have again 10 ways (from 0 to 9 both inclusive) to fill the 2nd place. Had the question constrained us that digits can't be repeated, we would have had 9 ways to fill the 2nd place( as one of the digits would have been used to fill the 1st place)

3rd place: The 3rd place can be filled with letters only. The total number of letters which we have is 26 (from A to Z both inclusive). So there are 26 ways in which we can fill the 3rd place.

4th place: The 4th place also needs to be filled with letters. Since the letters can be repeated, we have again 26 ways (from A to Z both inclusive) to fill the 4th place. Had the question constrained us that letters can't be repeated, we would have had 25 ways to fill the 4th place( as one of the letters would have been used to fill the 3rd place).

As the code constitutes of four characters, the number of ways of filling the four places can be written as = 10 * 10 * 26 * 26 = 67,600 ways.

Hope it's clear. Let me know if you have trouble at any point of this solution

Regards
Harsh

Regards Harsh

EMPOWERgmatRichC · Answer

Hi All

There are a couple of reasons why using the Combination Formula is incorrect:

1) We're dealing with unique codes, so order matters (the prompt states "...the same characters used in a different order constitute a different code..." Thus, permutation "math" is appropriate here.

2) Duplicate characters ARE allowed, so choosing one character does NOT impact how we choose the next.

When trying to decide whether to use Combination "math" or Permutation "math", it's usually best to do a quick 'sketch' of what you're after. If ABC is different from BAC and CBA, then it's a permutation. If a GROUP of letters (A, B and C) is the same group as (B, A and C), then it's a combination.

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

napagi · Answer

Great explanation of the governing concept. I'd like to add something, though. Since time is a concern I found myself using this shortcut: Instead of 10 x 10 x 26 x 26 I used 10 x 10 x 26 x 25. Changing the last 26 to a 25 allowed me to multiply by 100 then divide by 4 (=25) instead of multiplying by another 26. 10 x 10 x 26 x 100 / 4 10 x 10 x 26 = 2600, 2600 x 100 = 260.000, 260.000 / 4 = 65.000 You know that 26 is barely above 25, so you're looking for the answer that is above, but close to 65.000. 67.600 is the clear answer. Saved myself a good 20-30 seconds because I could do the whole calc in my head. Another look at it shows me a quicker way, as soon as you know that its 10 x 10 x 26 x 26 you know that the answer's last non-zero digit will be a 6 (you're first non-zero calculation in long multiplication will be 6 x 6 ). There's only one choice! Just my $0.02

.

neeraj609 · Answer

Do we need to worry about numbers repeating and coming up with duplicate combination?

thanks in advance!

ScottTargetTestPrep · Answer

We need to create a 4-digit code in which the first two characters are digits between 0 and 9 inclusive and the last two are selected from the 26 letters of the alphabet. Since characters can be repeated, we have:

Character 1 = 10 options

Character 2 = 10 options

Character 3 = 26 options

Character 4 = 26 options

Thus, the codes can be created in 10 x 10 x 26 x 26 = 67,600 ways.

Answer: B

ccooley · Answer

That's a great question.  In this problem, you don't need to worry about that.  Here's why.

Imagine that instead of what it says in the problem, you had just a two-digit code, and both digits had to be 1, 2, or 3. (I'm just doing this so that we get a comparable example that's smaller and easier to talk about.) As in the problem, digits can be repeated, and the same digits in a different order are different codes.

You'd multiply 3*3, and that wouldn't result in counting any cases twice. There are 9 possibilities.

The reason that works, is because when you multiply 3*3, you're doing something similar to 'counting paths'. There are three possible 'paths' at the very beginning:

1

2

3

Then, for each of these paths, there are three  more  paths when you're choosing the second digit:

1 - 1
1 - 2
1 - 3

2 - 1
2 - 2
2 - 3

3 - 1
3 - 2
3 - 3

Notice that only one of those paths gives you, for instance, '11' as a code. That's how you know you haven't counted '11' twice (even though it has two of the same digit).

qdeathf · Answer

Why is zero not excluded from the first option, since zero stops the ID code from being a four-digit number?

EMPOWERgmatRichC · Answer

Hi qdeathf,

The prompt asks us to determine the total number of 4-CHARACTER codes (not 4-"digit" codes), so we have to consider each of the possible values that can occur in each of the 4 'spots' of the code.

Although the original prompt has a 'typo' in it, we're told that the first two characters are DIGITS (meaning 0-9, inclusive, so 10 options) and the last two characters are LETTERS (A-Z, inclusive, so 26 options). Duplicate characters ARE allowed, so we have....

10 possible options for the first character
10 possible options for the second character
26 possible options for the third character
26 possible options for the fourth character

As such, "0" could be the first character in the code.
 
GMAT assassins aren't born, they're made,
Rich

Tommy21 · Answer

Hello,

Thanks a lot for your explanation. Could you please explain why 10p2 * 26p2 would be wrong in this case?

Thanks

EMPOWERgmatRichC · Answer

Hi Tommy21,

The calculation that you've set up assumes that the first and second characters CANNOT be the same and the third and fourth characters CANNOT be the same. However, the prompt tells us that characters can be "repeated, so the actual calculation would be:

(10)(10)(26)(26)
 
GMAT assassins aren't born, they're made,
Rich

Aannie · Answer

Could someone please explain , what does " same characters used in a different order constitute a different code" mean in the question then?

EMPOWERgmatRichC · Answer

Hi Aannie,

The prompt asks us to determine the total number of 4-CHARACTER codes (not 4-"digit" codes), so we have to consider each of the possible values that can occur in each of the 4 'spots' of the code. Although the original prompt has a 'typo' in it, we're told that the first two characters are DIGITS (meaning 0-9, inclusive, so 10 options) and the last two characters are LETTERS (A-Z, inclusive, so 26 options). Duplicate characters ARE allowed.

The prompt states that "the same characters used in a different order constitute a different code." For example...

12AB and 21BA are NOT the same code; they're different codes (so they both must be counted towards the total).

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

760+: What GMAT Assassins Do to Score at the Highest Levels

ThatDudeKnows · Answer

10*10*26*26

26*26 ends in a 6 and then we tack two zeros on in the hundreds and units digits.

Answer choice B.

AnushkaKala · Answer

how is the answer not 135.200 if the question mentions that arrangement can differ? - If characters may be repeated and the same characters used in a different order constitute a different code

Bunuel · Answer

The phrase “the same characters used in a different order constitute a different code” does not mean that the digits and letters can be placed anywhere.

The structure is fixed: the first two characters must be digits, and the last two must be letters.

So, for example, 12AB and 21AB are different codes, and 12AB and 12BA are different codes. This is already counted by doing:

10 * 10 * 26 * 26 = 67,600

The answer would become 135,200 only if both 12AB and AB12 were allowed. But AB12 is not allowed, because the first two characters must be digits.

kaustubhkohli12 · Answer

but then what exactky does it mean by characters may be repeated and same characters in different order. This is where i got confused as well .

Bunuel · Answer

It means two things:

Repetition is allowed, so codes such as 11AA or 00BB are valid. 
 Order matters within the fixed structure, so 12AB and 21AB are different, and 12AB and 12BA are different.

But the structure itself does not change: the first two characters must be digits, and the last two must be letters.

Each student at a certain university is given a four-charact

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