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How many four-digit odd numbers do not use any digit more than once?

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How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 09 Jun 2009, 08:31
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How many four-digit odd numbers do not use any digit more than once?

A. 1728
B. 2160
C. 2240
D. 2268
E. 2520

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.
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 09 Jun 2009, 10:05
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
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 09 Jun 2009, 17:00
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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.
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 09 Jun 2009, 22:45
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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
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 10 Jun 2009, 07:17
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..
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 10 Jun 2009, 09:33
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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.
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 05 Jan 2011, 08:21
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?
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 05 Jan 2011, 10:06
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.
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 05 Jan 2011, 12:34
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)
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 01 Jun 2011, 08:07
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?
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 01 Jun 2011, 08:25
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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"
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 01 Jun 2011, 16:42
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.
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 05 Nov 2015, 12:56
Just curious how do you know where the option for zero can be counted as well? i.e hundreds vs thousands? I would assume that as long as it is not counted in the unit digits it's all that matters.

It's quite clear to me why not the digits because then for example 7530 becomes divisible by 2. But what difference does 7053 and 7503 make?

Thank you!
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 23 Jul 2017, 04:16
EnergySP wrote:
How many four-digit odd numbers do not use any digit more than once?

A. 1728
B. 2160
C. 2240
D. 2268
E. 2520

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.


The question asks for 4 digit odd number which do not use any digit more than once.

So, the questions is basically asking to arrange the digits from 0,1,2,..,3,9 without repetition to form a 4 digit number with last digit as 1,3,5,7 or 9 only.

Lets say the 4 digit number be PQRS.
So, S can take numbers as 1,3,5,7,9 i.e. 5 ways.
Now we will chk nos. for P as it has restriction of being a non - zero. So it can take numbers as 1,2,..,9 - the no. taken in S which can be done in 8 ways,
Now, Q can take nos. 0,1,2,..., 9 minus the no. taken in S & P which can be done in 8 ways.
Now, R can take nos 0,1,2,...9 minus the no. taken in S, P & Q which can be done in 7 ways.

So, total count of such numbers = 8*8*7*5 = 2240

Answer C
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 31 Jul 2017, 06:45
bigoyal wrote:
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


I got the right answer but I'm not able to understand why this method is wrong.
please help Bunuel
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How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 03 Aug 2019, 01:29
How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Source: Math Bible (Jeff Sacksmann)
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 03 Aug 2019, 02:03
We have 5 options for the units digit (1,3,5,7 and 9)

We have 8 options for the thousands digit (1-9 except the digit used in the units place)

We have 8 options for the hundreds digit (0-9 except the digits used in units and thousands places)

We have 7 options for the tens digit (0-9 except the digits used in units, thousands and hundreds places)

So total number of four-digit odd numbers do not use any digit more than once = 8*8*7*5 = 2240

Answer is (C)

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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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New post 03 Aug 2019, 08:11
DarkHorse2019 wrote:
How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Source: Math Bible (Jeff Sacksmann)


given: there are total of 10 digits (0,1,2,3,4,5,6,7,8,9),
and odd numbers cannot have an even units digit, so there are 5 digits/choices (1,3,5,7,9).

units: 5 choices (because odd numbers cannot have even units digits)
thousands: 8 choices (10-1-1=8 choices because you can't have 0 and we used a digit for the units)
hundreds: 8 choices (10-2=8 digits left, since we used two digits for the thousands and units)
tens: 7 choices (10-3=7 digits left, since we used three digits for the thousands, hundreds and units)

total four-digit odd integers with different digits: 5*8*8*7=35*64=2240.

Answer (C).
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Re: How many four-digit odd numbers do not use any digit more than once?  [#permalink]

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Re: How many four-digit odd numbers do not use any digit more than once?   [#permalink] 03 Aug 2019, 11:16
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