M25 # 20 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 04:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M25 # 20

Author Message
Intern
Joined: 06 Jul 2009
Posts: 27
Followers: 0

Kudos [?]: 3 [0], given: 1

### Show Tags

10 Aug 2009, 14:53
Can someone give me a hand with understanding the formula behind this better?

I can't seem to wrap my head around it?

Thanks

How many odd three-digit integers greater than 800 are there such that all their digits are different?

(C) 2008 GMAT Club - m25#20

40
56
72
81
104
If the number begins with 8, there are possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit). If the number begins with 9, there are possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit). In all, there are numbers that satisfy the constraints.

Manager
Joined: 10 Jul 2009
Posts: 169
Followers: 1

Kudos [?]: 82 [0], given: 8

### Show Tags

10 Aug 2009, 15:30
Number of odd digit numbers starting with 8.
Hundreds place is filled with 8.
For a number to be odd, units place should be filled with either 1, 3, 5, 7, 9.
So there are 5 possibilities.
Tens place can be filled with either 0,1,2,3,4,5,6,7,9 .
But we have to eliminate the digits that are used in units position.
So number of odd numbers with unique digits starting with 8 = 1*8*5 = 40 ways.

For numbers starting with 9, unit digit can be either or 1,3,5,7
So number of odd numbers with unique digits starting with 9 = 1*8*4 = 32 ways.

So total number of possible combinations = 40 + 32 = 72 ways
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14457
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3724

Kudos [?]: 23029 [0], given: 4514

### Show Tags

10 Aug 2009, 17:03

Posted from my mobile device
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

Re: M25 # 20   [#permalink] 10 Aug 2009, 17:03
Similar topics Replies Last post
Similar
Topics:
46 M25#20 18 14 Mar 2009, 09:06
1 m25 - 20 8 30 Nov 2008, 05:20
18 m25#02 29 17 Oct 2008, 13:28
27 m25 #22 21 20 Oct 2008, 14:27
17 M25 #29 23 09 Oct 2008, 06:34
Display posts from previous: Sort by

# M25 # 20

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.