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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
A 3-character alpha-numeric code does have the following properties  [#permalink]

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Difficulty:   15% (low)

Question Stats: 84% (01:43) correct 16% (01:30) wrong based on 110 sessions

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A 3-character alpha-numeric code does have the following properties – the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?

A. 26
B. 64
C. 520
D. 6144
E. 9360

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A 3-character alpha-numeric code does have the following properties  [#permalink]

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2
EgmatQuantExpert wrote:
A 3-character alpha-numeric code does have the following properties – the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?

A. 26
B. 64
C. 520
D. 6144
E. 9360

The 3 character alpha-numeric code has the following characters

First character - Any number except 0 and 9 - Total of 10 - 2 = 8 possibilities
Second character - Any small letter except a and z - Total of 26 - 2 = 24 possibilities
Third character - Any number except 0 and 9 or Any small letter except a and z - Total of 8 + 24 = 32 possibilities

In this case, to find the total number of possibilities, we can use the unit's digit to find the
answer as number ending 8 * number ending 4 * number ending 2 is a number ending 4.
Also, the answer must be close to 10*20*30 is a little more than 6000.
P.S when the answer options are far apart, we can use approximations to arrive at the answer

Therefore, Option D(6144) is the the total number of 3-character alpha-numberic codes.
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Director  V
Joined: 12 Feb 2015
Posts: 956
Re: A 3-character alpha-numeric code does have the following properties  [#permalink]

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pushpitkc wrote:
EgmatQuantExpert wrote:
A 3-character alpha-numeric code does have the following properties – the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?

A. 26
B. 64
C. 520
D. 6144
E. 9360

The 3 character alpha-numeric code has the following characters

First character - Any number except 0 and 9 - Total of 10 - 2 = 8 possibilities
Second character - Any small letter except a and z - Total of 26 - 2 = 24 possibilities
Third character - Any number except 0 and 9 or Any small letter except a and z - Total of 8 + 24 = 32 possibilities

In this case, to find the total number of possibilities, we can use the unit's digit to find the
answer as number ending 8 * number ending 4 * number ending 2 is a number ending 4.
Also, the answer must be close to 10*20*30 is a little more than 6000.
P.S when the answer options are far apart, we can use approximations to arrive at the answer

Therefore, Option C(6144) is the the total number of 3-character alpha-numberic codes.

I think your calculations and answer is correct but by mistake you have mentioned option C in your answer (in brackets though the actual answer is right). I hope this helps.
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Manish "Only I can change my life. No one can do it for me"
Senior PS Moderator V
Joined: 26 Feb 2016
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Location: India
GPA: 3.12
Re: A 3-character alpha-numeric code does have the following properties  [#permalink]

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Once again, thanks Manish!
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Director  V
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Re: A 3-character alpha-numeric code does have the following properties  [#permalink]

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pushpitkc wrote:
Once again, thanks Manish!

You are welcome!! The short cuts of approximation and using the units digits are nice. Time saving.
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Manish "Only I can change my life. No one can do it for me"
e-GMAT Representative V
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Re: A 3-character alpha-numeric code does have the following properties  [#permalink]

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Solution

Given:
• The code is of 3-character alpha-numeric
• 1st character can be any number except 0 and 9
• 2nd character can be any small letter between a to z, excluded
• 3rd character can be any of those characters possible for the first two places

To find:
• Keeping all the constraints satisfied, how many such codes can be formed

Approach and Working:
• As the 1st character can be any number except 0 and 9, number of possibilities exist for the first character = 8
• As the 2nd character can be any small letter between a to z, excluded, number of possibilities exist for the second character = 24
• As the 3rd character can be any of those characters possible for the first two places, number of possibilities exist for the third character = 8 + 24 = 32
• Therefore, the total number of possible codes = 8 * 24 * 32 = 6144

Hence, the correct answer is option D.

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Re: A 3-character alpha-numeric code does have the following properties  [#permalink]

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Three different characters: C1 C2 C3
Theory: When calculating the number of items it is (xn-xa)+1

C1
8-1+1=8

C2
25th letter is y; 2nd letter is b
25-2+1=24

C3
8+24=32

Theory: Fundamental counting rule C1·C2·C3
8·24·32

I don't like multiplying 3 numbers together. I split the 8 up
(4·24)·(2·32)=96·64

Theory: the units digits of the answer results only from the units digits of the inputs
6·4=x4, so we know the answer ends in 4

B is too small. D must be the answer Re: A 3-character alpha-numeric code does have the following properties   [#permalink] 23 Dec 2018, 16:19
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