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

00:00

A

B

C

D

E

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.

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.

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

Answer should be D.

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 _________________

Re: Permutation question [#permalink]
09 Jun 2009, 16:00

1

This post received KUDOS

Thank you all who responded.

The answer is C, 2240.

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.

Re: Permutation question [#permalink]
09 Jun 2009, 21:45

1

This post received 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

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..

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. _________________

Re: Permutation question [#permalink]
05 Jan 2011, 07:21

EnergySP wrote:

Thank you all who responded.

The answer is C, 2240.

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?

Re: Permutation question [#permalink]
05 Jan 2011, 09:06

Expert's post

medanova wrote:

EnergySP wrote:

Thank you all who responded.

The answer is C, 2240.

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. _________________

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.

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?

Re: Permutation question [#permalink]
01 Jun 2011, 07:25

1

This post received 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)

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)