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Each student at a certain university is given a four-charact [#permalink]
01 May 2013, 08:47

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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]
01 May 2013, 09:01

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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 _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

Each student at a certain university is given a four-char [#permalink]
29 Sep 2013, 23:36

Expert's post

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-char [#permalink]
29 Sep 2013, 23:41

2

This post received KUDOS

Expert's post

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?

Each student at a certain university is given a four-charact [#permalink]
23 Aug 2014, 09:37

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

Re: Each student at a certain university is given a four-charact [#permalink]
26 May 2015, 09:32

Expert's post

Hi All

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.

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