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Tough probability question

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Tough probability question

by jfranco23 » Mon Jan 26, 2009 6:17 pm
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 code number. How many unassigned code numbers are there?

A. 6
B. 58
C. 174
D. 182
E. 399
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by gaggleofgirls » Mon Jan 26, 2009 6:57 pm
There are 8 choices (2-9 inclusive) for 3 spots, but once a choice is used, it cannot be used again.

For the first spot, we have 8 choices
For the second spot, we have 7 choices
For the third spot, we have 6 choices

8*7*6 = 336.

But the question asks how many unassigned codes there are after a code is assigned to each of the 330 employees.
336-330 = 6.

Answer = A.

-Carrie
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by krisraam » Mon Jan 26, 2009 6:58 pm
Answer is A.

No. of 3 digit numbers that can be formed with out repetition is
8 * 7 * 6 = 336

First digit can be any number from available 8 numbers. Second digit can be any number from the remaining 7 numbers and so on

Un assigned codes = 336 -330 = 6
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