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# How many 3 digit positive integers with distinct digits

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How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 00:09
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65% (hard)

Question Stats:

62% (02:57) correct 38% (03:07) wrong based on 151 sessions

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How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348
[Reveal] Spoiler: OA
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 00:21
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aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348

A number not to be a multiple of 10 should not have the units digit of 0.

XXX

9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)

9*8*8=576.

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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 00:59
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348

A number not to be a multiple of 10 should not have the units digit of 0.

XXX

9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)

9*8*8=576.

Hi Bunuel,

why is this solution wrong?
9C1 options for the first digit (from 1 to 9 inclusive).
9C1 options for the Second digit (from 0 to 9 inclusive minus the one we used for the first digit).
7C1 options for the third digit (from 1 to 9 inclusive minus 2 digits we used for the first and the second digits)

Which gives
9*9*7=567

Where i am wrong???

Rrsnathan.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 01:02
rrsnathan wrote:
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348

A number not to be a multiple of 10 should not have the units digit of 0.

XXX

9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)

9*8*8=576.

Hi Bunuel,

why is this solution wrong?
9C1 options for the first digit (from 1 to 9 inclusive).
9C1 options for the Second digit (from 0 to 9 inclusive minus the one we used for the first digit).
7C1 options for the third digit (from 1 to 9 inclusive minus 2 digits we used for the first and the second digits)

Which gives
9*9*7=567

Where i am wrong???

Rrsnathan.

If the second digit is 0, then you'll have 8 not 7 options for the third digit.

Hope it's clear.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 01:46
3
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Perfectly clear!! Thank you very much!
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Sep 2013, 10:21
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348

A number not to be a multiple of 10 should not have the units digit of 0.

XXX

9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)

9*8*8=576.

Hi Bunuel,

why is this solution wrong?
9C1 options for the first digit (from 1 to 9 inclusive).
9C1 options for the Second digit (from 0 to 9 inclusive minus the one we used for the first digit).
7C1 options for the third digit (from 1 to 9 inclusive minus 2 digits we used for the first and the second digits)

Which gives
9*9*7=567

Where i am wrong???

Rrsnathan.

If the second digit is 0, then you'll have 8 not 7 options for the third digit.

Hope it's clear.

It will still be wrong it is 9*9*8 and not 9*8*8...
still confused
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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17 Apr 2014, 04:26
Shouldn't it be 9x9x8?
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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17 Apr 2014, 04:32
pretzel wrote:
Shouldn't it be 9x9x8?

Nope. Check here: how-many-3-digit-positive-integers-with-distinct-digits-159249.html#p1264268
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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20 Apr 2014, 22:07
I did in this way :

case 1: tens place is 0 : 8 x 1 x 9 = 72

case 2 : tens place is not 0 : 7 x 8 x9 = 504

therefore , total numbers = 72 + 504 = 576 ans.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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05 Dec 2014, 19:01
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?

Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?

A. 576
B. 520
C. 504
D. 432
E. 348

A number not to be a multiple of 10 should not have the units digit of 0.

XXX

9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)

9*8*8=576.

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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Dec 2014, 12:05
Amit0507 wrote:
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?

hi, there can be certain cases in which zero can occur at the 3rd digit, if one follows 1st-2nd-3rd digit option. in order to eliminate all such cases we go with 1st-3rd-2nd digit option.

remember a number will be divisible by 10 if its last digit ends with zero, and in this case we want to eliminate all such cases in which zero occurs at the unit digit.

i hope it helps.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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06 Dec 2014, 18:06
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.

manpreetsingh86 wrote:
Amit0507 wrote:
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?

hi, there can be certain cases in which zero can occur at the 3rd digit, if one follows 1st-2nd-3rd digit option. in order to eliminate all such cases we go with 1st-3rd-2nd digit option.

remember a number will be divisible by 10 if its last digit ends with zero, and in this case we want to eliminate all such cases in which zero occurs at the unit digit.

i hope it helps.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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07 Dec 2014, 01:57
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.

lets try to understand, why we approach this question from first digit and why not from second or third digit ??

we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_

now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8

so the number is 98_

now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.

hence we have to eliminate all such possible cases.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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07 Dec 2014, 02:44
You mixed possibilities with the number at the highlighted section. Let me explain my approach.
1. I have 9 possibilities for the 1st digit.
2. I have again 9 possibilities for the second place as I will consider 0 here
3. Now out of 10, 2 numbers are used. If I assume 0 was used in the second place (10_) then I have 8 possibilities for the last place as well. so the final possibility will be 9*9*8.
I hope I made myself clear on the understanding. Please point out the flaw.

manpreetsingh86 wrote:
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.

lets try to understand, why we approach this question from first digit and why not from second or third digit ??

we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_

now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8

so the number is 98_

now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.

hence we have to eliminate all such possible cases.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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07 Dec 2014, 02:57
Amit0507 wrote:
You mixed possibilities with the number at the highlighted section. Let me explain my approach.
1. I have 9 possibilities for the 1st digit.
2. I have again 9 possibilities for the second place as I will consider 0 here
3. Now out of 10, 2 numbers are used. If I assume 0 was used in the second place (10_) then I have 8 possibilities for the last place as well. so the final possibility will be 9*9*8.
I hope I made myself clear on the understanding. Please point out the flaw.

manpreetsingh86 wrote:
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.

lets try to understand, why we approach this question from first digit and why not from second or third digit ??

we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_

now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8

so the number is 98_

now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.

hence we have to eliminate all such possible cases.

how do you know that zero will be at the second position. ?? i have highlighted one of the cases in which zero will not be at the second position.

and my dear friend, i'm not mixing any possibilities. that's the correct way of approaching the problem.
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Re: How many 3 digit positive integers with distinct digits [#permalink]

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07 Feb 2016, 09:57
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Re: How many 3 digit positive integers with distinct digits   [#permalink] 07 Feb 2016, 09:57
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