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Each student at a certain university is given a four-charact [#permalink]

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01 May 2013, 09:47

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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

Re: Each student at a certain university is given a four-charact [#permalink]

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01 May 2013, 10:01

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manishuol wrote:

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

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
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Each student at a certain university is given a four-char [#permalink]

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30 Sep 2013, 00:36

Source : Jeff Sackmann Extreme Challenge

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?

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?

Each student at a certain university is given a four-charact [#permalink]

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23 Aug 2014, 10:37

karishmatandon wrote:

manishuol wrote:

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

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

Can't it be Letter.Number.Letter.Number? Doesn't this add further combinations?

Additional question -- I followed the approach of 26C2 * 10C2. Why is that wrong? Is it because the combination formula take's the order into account? How would I unorder it?

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

Although the question is from last year, it doesn't look like anyone answered russ9's question.

There are a couple of reasons why using the Combination Formula (as he did) 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.

Each student at a certain university is given a four-charact [#permalink]

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14 Dec 2015, 09:17

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karishmatandon wrote:

manishuol wrote:

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

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

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 [= 36]). There's only one choice!

Re: Each student at a certain university is given a four-charact [#permalink]

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14 Dec 2015, 15:52

Bunuel...Why are we not multiplying it by 4! to account for the different ordering that can be got.

Bunuel wrote:

GNPTH wrote:

Source : Jeff Sackmann Extreme Challenge

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?

Re: Each student at a certain university is given a four-charact [#permalink]

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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

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

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

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
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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

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

thanks in advance!

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).
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Re: Each student at a certain university is given a four-charact [#permalink]

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02 Sep 2017, 11:52

Avinashs87 wrote:

Bunuel...Why are we not multiplying it by 4! to account for the different ordering that can be got.

To anyone wondering why, multiplying by 4! means getting different orders like Number.Number.Alphabet.Alphabet and Number.Alphabet.Number.Alphabet which is not required here.