Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 21 Aug 2014, 12:43

### 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

# Permutation question

Author Message
TAGS:
Intern
Joined: 20 Jul 2008
Posts: 4
Followers: 0

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

Permutation question [#permalink]  09 Jun 2009, 07:31
00:00

Difficulty:

(N/A)

Question Stats:

38% (02:50) correct 62% (00:39) wrong based on 25 sessions
Q. How many four-digit odd numbers do not use any digit more than once?
A. 1728;
B. 2160;
C. 2240;
D. 2268;
E 2520

Please provide your explanation when answering the question. I have an answer in front of me, along with the explanation. But, the explanation is not clear ... looking for an alternative explanation. Thank you.

[Reveal] Spoiler:

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.
Founder
Affiliations: UA-1K, SPG-G, HH-D
Joined: 04 Dec 2002
Posts: 12065
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2219

Kudos [?]: 8200 [0], given: 3641

Re: Permutation question [#permalink]  09 Jun 2009, 07:35
Expert's post
Please continue discussion in the PS Forum
_________________

Founder of GMAT Club

Just starting out with GMAT? Start here... | Want to know your GMAT Score? Try GMAT Score Estimator
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Have a blog? Feature it on GMAT Club!

Director
Joined: 03 Jun 2009
Posts: 805
Location: New Delhi
WE 1: 5.5 yrs in IT
Followers: 62

Kudos [?]: 386 [0], given: 56

Re: Permutation question [#permalink]  09 Jun 2009, 09:05

Let a 4-digit number be represented by ABCD

Here A can have any value between 1 to 9 - so total 9
B can have any value between 0 to 9, but not A - so total 9
C can have any value between 0 to 9, but not A or B - so total 8
D can have any value between 0 to 9, but not A, B or C - so total 7

No. of ALL possible 4-digit nos (without repeating any digit) = 9*9*8*7 = 4536
Half of these would be odd.
Therefor, no. of ODD possible 4-digit nos (without repeating any digit) = 4536 / 2 = 2268
_________________
Intern
Joined: 04 Mar 2008
Posts: 44
Followers: 0

Kudos [?]: 3 [0], given: 6

Re: Permutation question [#permalink]  09 Jun 2009, 09:09
The asnwer is E, 2520.
Director
Joined: 03 Jun 2009
Posts: 805
Location: New Delhi
WE 1: 5.5 yrs in IT
Followers: 62

Kudos [?]: 386 [0], given: 56

Re: Permutation question [#permalink]  09 Jun 2009, 11:18
abhitheCEO wrote:
The asnwer is E, 2520.

will appreciate if you can provide the details as well.
_________________
Intern
Joined: 27 Apr 2009
Posts: 12
Followers: 0

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

Re: Permutation question [#permalink]  09 Jun 2009, 15:46
my choice is 'B' 2160.
OA plz.
Intern
Joined: 20 Jul 2008
Posts: 4
Followers: 0

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

Re: Permutation question [#permalink]  09 Jun 2009, 16:00
1
KUDOS
Thank you all who responded.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.
Director
Joined: 03 Jun 2009
Posts: 805
Location: New Delhi
WE 1: 5.5 yrs in IT
Followers: 62

Kudos [?]: 386 [0], given: 56

Re: Permutation question [#permalink]  09 Jun 2009, 21:30
Got it. Now I understand where I was wrong in my calculation .
_________________
Intern
Joined: 12 May 2009
Posts: 48
Location: Mumbai
Followers: 1

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

Re: Permutation question [#permalink]  09 Jun 2009, 21:45
1
KUDOS
This question has to be solved reverse
ABCD is a 4 digit number

Since D (the unit digit) has to be odd, we have 5 choices => D=5
Starting from A;
A can hold 1 to 9, but cannot contain the digit in unit position, so choices will be 8 => A = 8
B can gold 0 to 9 (10 choices), but not the digits in A and D, so we ahve 8 options => B = 8
C can also hold 0 to 9 (10 choices), but not the digits in A, B and D, se we have 7 choices => C = 7

Calculating the total number of options = A*B*C*D = 8*8*7*5 = 2240

IMO C
Manager
Joined: 28 Jan 2004
Posts: 204
Location: India
Followers: 2

Kudos [?]: 10 [0], given: 4

Re: Permutation question [#permalink]  09 Jun 2009, 23:25
2240.
Exactlt same sol. as EnergySP.
Senior Manager
Joined: 15 Jan 2008
Posts: 295
Followers: 2

Kudos [?]: 22 [0], given: 3

Re: Permutation question [#permalink]  10 Jun 2009, 06:17
I wonder as to why shud i not solve it the other around...

tha last digit can have 5 ways.
the second last digit can have 9 ways. ( 0-9) except the last digit.
the third last digit cab have 8 ways ( 0-9) except the last and the second last digit.
the fourth last digit can have 6 ways ( as no zero, last digiht, second last, thirdt last)

hence the total will be 45*48 = 2160 ways..

can anyone tell me from where am i missing the extra 80 ways..
CEO
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 355

Kudos [?]: 1736 [0], given: 357

Re: Permutation question [#permalink]  10 Jun 2009, 08:33
Expert's post
Neochronic wrote:
I wonder as to why shud i not solve it the other around...

tha last digit can have 5 ways.
the second last digit can have 9 ways. ( 0-9) except the last digit.
the third last digit cab have 8 ways ( 0-9) except the last and the second last digit.
the fourth last digit can have 6 ways ( as no zero, last digiht, second last, thirdt last)

hence the total will be 45*48 = 2160 ways..

can anyone tell me from where am i missing the extra 80 ways..

The highlighted part is correct only if second, third and fourth digits don't equal zero. Otherwise, we will have 7 ways.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 19 Aug 2010
Posts: 78
Followers: 2

Kudos [?]: 9 [0], given: 2

Re: Permutation question [#permalink]  05 Jan 2011, 07:21
EnergySP wrote:
Thank you all who responded.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.

I know it is an old post but I received 2160 as an answer and can´t figure out which one is correct.

I start from the last digit:
5 possiblities for the last digit
9 possibilities or the 3rd digit
8 for the 2nd and
6 for the 1st

So we have 6*8*9*5=2160

But EnergySP solution is also correct. Now I am confused.
Can somebody help?
Math Expert
Joined: 02 Sep 2009
Posts: 19038
Followers: 3362

Kudos [?]: 24464 [0], given: 2677

Re: Permutation question [#permalink]  05 Jan 2011, 09:06
Expert's post
medanova wrote:
EnergySP wrote:
Thank you all who responded.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.

I know it is an old post but I received 2160 as an answer and can´t figure out which one is correct.

I start from the last digit:
5 possiblities for the last digit
9 possibilities or the 3rd digit
8 for the 2nd and
6 for the 1st

So we have 6*8*9*5=2160

But EnergySP solution is also correct. Now I am confused.
Can somebody help?

See Walker's post above: you'll have 6 choices for the 1st digit if 2nd or 3rd digit doesn't equal to zero, otherwise you'll have 7 choices. So this approach is not correct.
_________________
Intern
Joined: 21 Sep 2010
Posts: 17
Followers: 0

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

Re: Permutation question [#permalink]  05 Jan 2011, 11:34
four Digit Number which is odd i.e it must end with 1,3,5,7 or 9

So if following 4 blanks represent the 4 digit number:

- - - -

Unit's place can be filled in by any of the 5 digits (1,3,5,7 or9)

so we get,

- - - 5

We are left with 9 other numbers to fill in the rest, but we cannot repeat and first digit cannot be zero otherwise the number will not be truly 4 digit number.

Hence,

8*8*7*5= 2240 (ans. C)
Intern
Joined: 12 May 2011
Posts: 2
Followers: 0

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

Re: Permutation question [#permalink]  01 Jun 2011, 07:07
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2048
Followers: 128

Kudos [?]: 898 [1] , given: 376

Re: Permutation question [#permalink]  01 Jun 2011, 07:25
1
KUDOS
CyberAsh wrote:
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?

I think you should look for something called "Slot Method" in MGMAT guide for P&C.

The idea is to assign the number in ascending order of restriction.

Units place: 1,3,5,7,9- Total=5(Most restrictive)
Thousands place: No 0 and not the digit used by units place- Total=8 (Less restrictive)
Tens place: 2 digits used. Left:10-2=8 (Yet less restrictive)
Hundreds place: 3 digits used. Left:10-3=7(Yet less restrictive)

Total=5*8*8*7=2240

Ans: "C"
_________________
Intern
Joined: 12 May 2011
Posts: 2
Followers: 0

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

Re: Permutation question [#permalink]  01 Jun 2011, 15:42
fluke wrote:
CyberAsh wrote:
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?

I think you should look for something called "Slot Method" in MGMAT guide for P&C.

The idea is to assign the number in ascending order of restriction.

Units place: 1,3,5,7,9- Total=5(Most restrictive)
Thousands place: No 0 and not the digit used by units place- Total=8 (Less restrictive)
Tens place: 2 digits used. Left:10-2=8 (Yet less restrictive)
Hundreds place: 3 digits used. Left:10-3=7(Yet less restrictive)

Total=5*8*8*7=2240

Ans: "C"
Thanks, much appreciated.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1365
Followers: 11

Kudos [?]: 135 [0], given: 10

Re: Permutation question [#permalink]  13 Jun 2011, 00:23
ABCD

D has 5 options
A has 8 options
B has 7 options + 1 option of 0.
C has 7 options

8*8*7*5
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Re: Permutation question   [#permalink] 13 Jun 2011, 00:23
Similar topics Replies Last post
Similar
Topics:
Permutations Question 0 05 Jul 2013, 08:35
1 Intermediate Permutation method/question 3 14 Oct 2008, 15:30
3 Permutation/Combination question 11 26 Sep 2008, 22:41
Permutation/ Combination Question 12 10 Dec 2006, 20:37
Permutation - Question Paper 1 05 Jan 2005, 22:42
Display posts from previous: Sort by