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permutations

by crackinggmat » Tue Oct 11, 2011 10:08 am
Pls help me find my mistake.....

Will must choose a 3 character computer password consisting of 1 letter from the alphabet and 2 distinct digits ,in any order. From how many different passwords can will choose??

A 390
B 2340
C 4680
D 7020
E 14040

My Solution

will can choose from 26 alphabets in 26 ways
he can choose 1st digit in 10 ways and other digit in 9 ways

therefore total number of ways = 26 * 10 * 9

Now he can arrange this combination of letter and digits in 6 ways

hence final answer should is 6* 26*10*9 = 14040

but the actual ans is 7020
this is question from princeton reveiw math workout

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by shankar.ashwin » Tue Oct 11, 2011 10:12 am
26 for the alphabets.
10C2 for the digits - 45.

Now this is arranged in 3! ways,

So 26*45*6=7020.

You did 10*9 for numbers, which includes ordering and again multiplied that by 3!. So those would be duplicates.

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by [email protected] » Tue Oct 11, 2011 10:17 am
crackinggmat wrote:Pls help me find my mistake.....

Will must choose a 3 character computer password consisting of 1 letter from the alphabet and 2 distinct digits ,in any order. From how many different passwords can will choose??

A 390
B 2340
C 4680
D 7020
E 14040

My Solution

will can choose from 26 alphabets in 26 ways
he can choose 1st digit in 10 ways and other digit in 9 ways

therefore total number of ways = 26 * 10 * 9

Now he can arrange this combination of letter and digits in 6 ways

hence final answer should is 6* 26*10*9 = 14040

but the actual ans is 7020
this is question from princeton reveiw math workout
I think the answer is right.

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by leonswati » Tue Oct 11, 2011 10:55 am
crackinggmat wrote:Pls help me find my mistake.....

Will must choose a 3 character computer password consisting of 1 letter from the alphabet and 2 distinct digits ,in any order. From how many different passwords can will choose??

A 390
B 2340
C 4680
D 7020
E 14040

My Solution

will can choose from 26 alphabets in 26 ways
he can choose 1st digit in 10 ways and other digit in 9 ways

therefore total number of ways = 26 * 10 * 9

Now he can arrange this combination of letter and digits in 6 ways

hence final answer should is 6* 26*10*9 = 14040

but the actual ans is 7020
this is question from princeton reveiw math workout


Shankar has explained it really well.... If u still have a doubt look at it this way...

There are three ways to arrange the alphabets :
1) A D D

2) D A D

3) D D A

As u have already Grouped the digits like if there are 10 ways to select the first digit then are are 9 ways to select the second digit .So reordering them again will create duplicates. But as alphabets have not been ordered so there are three ways to position the Alphabets ...

26*3*10*9 = 7020

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by [email protected] » Tue Dec 19, 2017 7:21 am
crackinggmat wrote:Pls help me find my mistake.....

Will must choose a 3 character computer password consisting of 1 letter from the alphabet and 2 distinct digits ,in any order. From how many different passwords can will choose??

A 390
B 2340
C 4680
D 7020
E 14040
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

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Re: permutations

by [email protected] » Sun Jan 26, 2020 9:08 am
crackinggmat wrote:
Tue Oct 11, 2011 10:08 am
Will must choose a 3 character computer password consisting of 1 letter from the alphabet and 2 distinct digits ,in any order. From how many different passwords can will choose??

A 390
B 2340
C 4680
D 7020
E 14040
Take the task of creating a password and break it into stages.

Stage 1: Select the one letter to be used in the code
There are 26 letters from which to choose, so we can complete this stage in 26 ways.

Stage 2: Select the two digits to be used in the code
Since the order in which we select the two digits does not matter, we can use combinations.
We can select 2 digits from 10 women in 10C2 ways (45 ways)
So, we can complete stage 2 in 45 ways

NOTE: We now have the 3 characters to be used in the code. At this point, we need to arrange those 3 characters.

Stage 3: Arrange the 3 selected characters.
RULE: We can arrange n unique objects in n! ways.
So, we can arrange the 3 characters in 3! ways (6 ways)
So we can complete stage 3 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a password ) in (26)(45)(6) ways
(26)(45)(6) = 7020

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
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