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

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Manager
Joined: 03 Oct 2008
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A company has assigned a distinct 3-digit code number to [#permalink]

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06 Oct 2008, 07:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Senior Manager
Joined: 04 Jan 2006
Posts: 276

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06 Oct 2008, 12:49
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

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SVP
Joined: 29 Aug 2007
Posts: 2472

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06 Oct 2008, 14:59
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

A. exactly.

= 8c3 x 3!
=336
so the remaining = 336 - 330
= 6
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Senior Manager
Joined: 04 Aug 2008
Posts: 372

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06 Oct 2008, 15:08
isnt this a permutation?

3P8=8*7*6=336... etc
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Senior Manager
Joined: 21 Apr 2008
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06 Oct 2008, 15:52
Since the order matters (234 is different from 342, 423 etc)

8P3 - 300 = 6

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07 Oct 2008, 03:55
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.

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Re: math problem   [#permalink] 07 Oct 2008, 03:55
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