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Company X ordered for security codes to be formed [#permalink]
27 Nov 2019, 23:37

00:00

Question Stats:

14% (02:39) correct
85% (01:12) wrong based on 7 sessions

Company X ordered for security codes to be formed for each of its employees. Codes should be designed such that each code has five characters comprising of 3 letters and 2 digits. If only 3 digits (1,2,3) can be used and only 2 letters (X,Y) can be used for the codes, then how many different codes can be formed such that repetition of the letters and digits is allowed

Re: Company X ordered for security codes to be formed [#permalink]
28 Nov 2019, 08:53

1

This post received KUDOS

3 alphabets + 2 digits

3 alphabets out of (X, Y) Repetition is allowed.

3 alphabets options --> { 3x, 3y, 2x+y , x+2y } --> 4 options out of these 4 options alike option pairs are {3x, 3y} and { 2x+y, x+2y}

2 digits out of 3 options (1,2,3) The option for digits can be {11, 22, 33,12,13,23} alike options for digits {11, 22, 33} and {12,13,23}

Let's find all possible solutions now {3x, 3y} with {11, 22, 33} {3x, 3y} with {12,13,23} { 2x+y, x+2y} with {11, 22, 33} { 2x+y, x+2y} with {12,13,23}

option a {3x, 3y} with {11, 22, 33}

Note: 2-> no of terms in first set , 3-> no of terms in set 2 , (5!/(3!X2!) --> no of options when out of total 5 positions, 3 are repiting in set1 and 2 are repiting in set 2 For more clarity check the video permutation with repetition. https://www.ck12.org/probability/permut ... n-BSC-PST/

2X3X(5!/(3!X2!) ) = 60

option b {3x, 3y} with {12,13,23}

2X3X(5!/(3!)) = 120

option c { 2x+y, x+2y} with {11, 22, 33}

2X3X(5!/(2!X2!)) = 180

option d { 2x+y, x+2y} with {12,13,23}

2X3X(5!/(2!))v= 360

Adding all the option --> 720

Hence, D is the Answer.

The answer might seem tedious, but it is simple once you get hang of it. Please do let me know in case you have any queries.

Re: Company X ordered for security codes to be formed [#permalink]
28 Nov 2019, 10:14

Expert's post

RSQUANT wrote:

Company X ordered for security codes to be formed for each of its employees. Codes should be designed such that each code has five characters comprising of 3 letters and 2 digits. If only 3 digits (1,2,3) can be used and only 2 letters (X,Y) can be used for the codes, then how many different codes can be formed such that repetition of the letters and digits is allowed?

(a)72 (b)180 (c)360 (d)720 (e)1440

Here's a different approach...

Take the task of creating codes and break it into stages.

Stage 1: Determine the GENERIC arrangement of digits and letters. For example, one possible arrangement is DIGIT-LETTER-LETTER-DIGIT-LETTER Another possible arrangement is LETTER-DIGIT-DIGIT-LETTER-LETTER

Let L represent a letter Let D represent a digit So we want to arrange 3 L's and 2 D's

-----ASIDE------ When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows: There are 11 letters in total There are 4 identical I's There are 4 identical S's There are 2 identical P's So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)] --------------------------

We want to arrange 3 L's and 2 D's. So: There are 5 characters in total There are 3 identical L's There are 2 identical D's So, the total number of possible arrangements = 5!/[(3!)(2!)] = 10 So, we can complete stage 1 in 10 ways

Stage 2: Select a letter (X or Y) to replace the 1st L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete stage 2 in 2 ways

Stage 3: Select a letter (X or Y) to replace the 2nd L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete stage 3 in 2 ways

Stage 4: Select a letter (X or Y) to replace the 3rd L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete this stage in 2 ways

Stage 5: Select a digit (1, 2 or 3) to replace the 1st D in the arrangement (the arrangement from stage 1) We can choose 1, 2 or 3, so we can complete this stage in 3 ways

Stage 6: Select a digit (1, 2 or 3) to replace the 2nd D in the arrangement (the arrangement from stage 1) We can choose 1, 2 or 3, so we can complete this stage in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus create a code) in (10)(2)(2)(2)(3)(3) ways (= 720 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: Company X ordered for security codes to be formed [#permalink]
06 Dec 2019, 04:11

GreenlightTestPrep wrote:

RSQUANT wrote:

Company X ordered for security codes to be formed for each of its employees. Codes should be designed such that each code has five characters comprising of 3 letters and 2 digits. If only 3 digits (1,2,3) can be used and only 2 letters (X,Y) can be used for the codes, then how many different codes can be formed such that repetition of the letters and digits is allowed?

(a)72 (b)180 (c)360 (d)720 (e)1440

Here's a different approach...

Take the task of creating codes and break it into stages.

Stage 1: Determine the GENERIC arrangement of digits and letters. For example, one possible arrangement is DIGIT-LETTER-LETTER-DIGIT-LETTER Another possible arrangement is LETTER-DIGIT-DIGIT-LETTER-LETTER

Let L represent a letter Let D represent a digit So we want to arrange 3 L's and 2 D's

-----ASIDE------ When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows: There are 11 letters in total There are 4 identical I's There are 4 identical S's There are 2 identical P's So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)] --------------------------

We want to arrange 3 L's and 2 D's. So: There are 5 characters in total There are 3 identical L's There are 2 identical D's So, the total number of possible arrangements = 5!/[(3!)(2!)] = 10 So, we can complete stage 1 in 10 ways

Stage 2: Select a letter (X or Y) to replace the 1st L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete stage 2 in 2 ways

Stage 3: Select a letter (X or Y) to replace the 2nd L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete stage 3 in 2 ways

Stage 4: Select a letter (X or Y) to replace the 3rd L in the arrangement (the arrangement from stage 1) We can choose X or Y, so we can complete this stage in 2 ways

Stage 5: Select a digit (1, 2 or 3) to replace the 1st D in the arrangement (the arrangement from stage 1) We can choose 1, 2 or 3, so we can complete this stage in 3 ways

Stage 6: Select a digit (1, 2 or 3) to replace the 2nd D in the arrangement (the arrangement from stage 1) We can choose 1, 2 or 3, so we can complete this stage in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus create a code) in (10)(2)(2)(2)(3)(3) ways (= 720 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.