Q5: If a code word is defined to be a sequence of different : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 01:25

### 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

# Q5: If a code word is defined to be a sequence of different

Author Message
Manager
Joined: 11 Jan 2007
Posts: 197
Location: Bangkok
Followers: 1

Kudos [?]: 63 [0], given: 0

Q5: If a code word is defined to be a sequence of different [#permalink]

### Show Tags

05 Jun 2007, 19:26
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Q5:
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
_________________

cool

Director
Joined: 26 Feb 2006
Posts: 904
Followers: 4

Kudos [?]: 107 [0], given: 0

### Show Tags

05 Jun 2007, 19:54
jet1445 wrote:
Q5:If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1

E. without repetation
= (10 x 9 x 8 x 7 x 6) /(10 x 9 x 8 x 7) = 6:1

with repetation = (10^5)/ (10^4) = 10:1
Manager
Joined: 17 Oct 2006
Posts: 52
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

06 Jun 2007, 02:16
Its a permutation problem because the order does matter i.e., abcde is different than edcba. When we have decided that it is a permutation problem, then there is just simple math to do.
so from the data of the question
10P5 : 10P4
= 10!/(10-5)! : 10!/(10-4)!
= 10!/5! : 10!/6!
= 10*9*8*7*6*5!/5! :10*9*8*7*6!/6!
= 10*9*8*7*6 : 10*9*8*7
so 10*9*8*7*6/10*9*8*7
=6/1
=6:1 so E
Manager
Joined: 20 Dec 2004
Posts: 179
Followers: 2

Kudos [?]: 17 [0], given: 0

### Show Tags

06 Jun 2007, 15:20
It's a permutation problem
10P5/10P4 = 6:1
_________________

Regards

Subhen

06 Jun 2007, 15:20
Display posts from previous: Sort by