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A company has assigned a distinct 3-digit code number to

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A company has assigned a distinct 3-digit code number to [#permalink] New post 06 Oct 2008, 06:31
A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits

2, 3, 4, 5, 6, 7, 8, 9

and no digit appears more than once in any one code number. How many unassigned code numbers are there?

A. 6
B. 58
C. 174
D. 182
E. 399
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Re: math problem [#permalink] New post 06 Oct 2008, 11:49
albany09 wrote:
A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits

2, 3, 4, 5, 6, 7, 8, 9

and no digit appears more than once in any one code number. How many unassigned code numbers are there?

A. 6
B. 58
C. 174
D. 182
E. 399


Possible code number = P(8,3) = 8! / (8 - 3)! = 8 x 7 x 6 = 336 numbers

The number of unassigned code = 336 - 330 = 6

The answer is A.
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Re: math problem [#permalink] New post 06 Oct 2008, 13:59
devilmirror wrote:
albany09 wrote:
A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits

2, 3, 4, 5, 6, 7, 8, 9

and no digit appears more than once in any one code number. How many unassigned code numbers are there?

A. 6
B. 58
C. 174
D. 182
E. 399


Possible code number = P(8,3) = 8! / (8 - 3)! = 8 x 7 x 6 = 336 numbers

The number of unassigned code = 336 - 330 = 6

The answer is A.


A. exactly.

= 8c3 x 3!
=336
so the remaining = 336 - 330
= 6
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Re: math problem [#permalink] New post 06 Oct 2008, 14:08
isnt this a permutation?

3P8=8*7*6=336... etc
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Re: math problem [#permalink] New post 06 Oct 2008, 14:52
Since the order matters (234 is different from 342, 423 etc)

8P3 - 300 = 6
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Re: math problem [#permalink] New post 07 Oct 2008, 02:55
Irrespective of permutation of combination, my approach is the following:

hundred's place can be occupied by 8 digits. Thus, only 7 digits are remaining for 10's place and only 6 remaining for unit's place (as no digit can repeat).

Hence, total possible codes = 8*7*6 = 336 and subtracting 300 from this will give 6.
Re: math problem   [#permalink] 07 Oct 2008, 02:55
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