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Combinatorics problem

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by abhi75 » Thu May 01, 2008 7:42 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 one code number. How many unassigned code numbers are there?

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

Can someone please tell me how to approach this kind of problem.

Thanks.
Abhi
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by netigen » Thu May 01, 2008 8:26 pm
there are 8 numbers and no repetitions so lets say the number is XYZ

for X there are 8 possibilities
for Y there are 7
for Z there are 6

total possible numbers = 8 x 7 x 6 = 336

so the answer is 6
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by dextar » Thu May 01, 2008 8:27 pm
Hi

I think the answer should be 6 i.e A

See there are 8 choices to choose from 2 to 9 at hundrede's place

When u choose a digit, the no of choices remaining would be 7 for the tenth and 6 for the unit's digit.

So total chouces are
8*7*6=336

Since the total employees are 330, so unassigned numbers would be 336-330=6

Got it?
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