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

Expert replies
by imane81 » Wed Feb 03, 2010 1:48 pm
Your help on this one is much appreciated. Thanks in advance

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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Source: — Problem Solving |

by Ian Stewart » Wed Feb 03, 2010 2:42 pm
imane81 wrote:Your help on this one is much appreciated. Thanks in advance

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
To make a code number, we have 8 choices for the first digit, 7 choices for the second digit, and 6 choices for the third digit (subtracting one each time, since we cannot use the same digit more than once), and therefore 8*7*6 = 336 code numbers are possible in total (we just need to multiply the number of choices we have for each digit to find how many code numbers can be made). Since 330 code numbers have been used already, there are 336-330 = 6 unused code numbers.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by neelimareddym » Thu Feb 04, 2010 4:43 am
Total number of codes that can be formed = 8*7*6 =336
Therefore, Number left unused = 336 - 330 = 6
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