Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 04:48

### 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

# Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit

Author Message
Manager
Joined: 14 Mar 2007
Posts: 235
Followers: 1

Kudos [?]: 7 [0], given: 0

Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit [#permalink]

### Show Tags

18 Apr 2007, 06:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit numbers, in which at least 2 digits are the same. Therefore, x must be:

(A) 505
(B) 427
(C) 120
(D) 625
(E) 384
Manager
Joined: 02 Jan 2007
Posts: 208
Followers: 2

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

### Show Tags

18 Apr 2007, 06:34
we can do this if we can find out 2 things:
1) The number of 4-digit numbers where none of the digits are the same.
2) The total number of 4-digits numbers that we can form with the given numbers.

2) - 1) will give us the answer.

To get 1)
5*4*3*2 = 120

To get 2)
5*5*5*5 = 625

thta gives 505 .....(A)
Manager
Joined: 14 Mar 2007
Posts: 235
Followers: 1

Kudos [?]: 7 [0], given: 0

### Show Tags

18 Apr 2007, 06:44
correct and nice explanation
Manager
Joined: 28 Feb 2007
Posts: 197
Location: California
Followers: 3

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

### Show Tags

18 Apr 2007, 21:26
andrehaui wrote:
Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit numbers, in which at least 2 digits are the same. Therefore, x must be:

(A) 505
(B) 427
(C) 120
(D) 625
(E) 384

There are 5 possible numbers.
Number of all possible 4 digit numbers = 5^4 = 625
Number of 4 digit numbers with distinct numbers = 5.4.3.2 = 120
Hence answer is 625 - 120 = 505

A
Re: combination   [#permalink] 18 Apr 2007, 21:26
Display posts from previous: Sort by