BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Combinatorics problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 105
Joined: Tue Oct 16, 2007 5:57 am
Thanked: 3 times

Combinatorics problem

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
Top
Source: — Problem Solving |
Legendary Member
Posts: 631
Joined: Mon Feb 18, 2008 11:57 pm
Thanked: 29 times
Followed by:3 members

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
Top
Master | Next Rank: 500 Posts
Posts: 100
Joined: Tue Jan 29, 2008 12:08 am
Location: INDIA
Thanked: 3 times
Followed by:1 members

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?
Top
Post new topic Post Reply
• Page 1 of 1