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
permutations
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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.
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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I think the answer is right.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
 leonswati
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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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Since there are 26 letters and 10 digits, the number of 3character 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.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
Answer: D
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Take the task of creating a password and break it into stages.crackinggmat wrote: ↑Tue Oct 11, 2011 10:08 amWill 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
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/whatshould ... 67256.html
 https://www.beatthegmat.com/countingpro ... 44302.html
 https://www.beatthegmat.com/pickinga5 ... 73110.html
 https://www.beatthegmat.com/permutation ... 57412.html
 https://www.beatthegmat.com/simpleonet270061.html
MEDIUM
 https://www.beatthegmat.com/combinatoric ... 73194.html
 https://www.beatthegmat.com/arabianhors ... 50703.html
 https://www.beatthegmat.com/subsetspro ... 73337.html
 https://www.beatthegmat.com/combinatoric ... 73180.html
 https://www.beatthegmat.com/digitsnumberst270127.html
 https://www.beatthegmat.com/doubtonsep ... 71047.html
 https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
 https://www.beatthegmat.com/wonderfulp ... 71001.html
 https://www.beatthegmat.com/permutation ... 73915.html
 https://www.beatthegmat.com/permutationt122873.html
 https://www.beatthegmat.com/notwoladie ... 75661.html
 https://www.beatthegmat.com/combinationst123249.html
Cheers,
Brent