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# How many numbers from 1 to 1000, both inclusive, have digits repeated

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Manager
Joined: 14 Feb 2018
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How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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Updated on: 14 Apr 2018, 10:31
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Question Stats:

65% (01:35) correct 35% (01:21) wrong based on 26 sessions

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How many numbers from 1 to 1000, both inclusive, have digits repeated ?

A. 9
B. 81
C. 262
D. 648
E. 738
[Reveal] Spoiler: OA

Originally posted by SonalSinha803 on 14 Apr 2018, 09:23.
Last edited by Bunuel on 14 Apr 2018, 10:31, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Posts: 5776
Re: How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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14 Apr 2018, 09:46
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SonalSinha803 wrote:
How many numbers from 1 to 1000, both inclusive, have digits repeated ?

A. 9
B. 81
C.262
D.648
E.738

Will post the OA and OE after a few discussions.

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Hi..

3-digits without repetition = 9*9*8=648
2-digits without repetition = 9*9=81
1-digit without repetition= 9

Total = 648+81+9=738..

So numbers with repetition = 1000-738=262
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How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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14 Apr 2018, 12:48
SonalSinha803 wrote:
How many numbers from 1 to 1000, both inclusive, have digits repeated ?

A. 9
B. 81
C. 262
D. 648
E. 738

1-99➡9 numbers
1000➡1 number
100-999➡252 numbers:
xxx=9*1*1=9
xxy=9*1*9=81
xyx=9*9*1=81
yxx=9*9*1=81
252+9+1=262 numbers
C
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Joined: 28 Mar 2018
Posts: 3
Re: How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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14 Apr 2018, 13:04
I don't get this. Can someone explain more? Somebody please. Thanks

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How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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15 Apr 2018, 00:15
eddwwaarrd wrote:
I don't get this. Can someone explain more? Somebody please. Thanks

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Hi eddwwaarrd

I have merely elaborated on solutions given by chetan2u and gracie

Between 1 and 1000(both inclusive),
there are 1 digit numbers, 2 digit numbers, and 3 digit numbers.

There are no 1 digit number that has repeated digits [9 possibilities]

For two digit numbers, 0 can't be the first digit
The first digit can be 1-9 [9 possibilities]
The second digit can be 0,1-9(except the first digit) [9 possibilities]
Total possible 2 digit numbers are 9*9 = 81

For three digit numbers, again 0 can't be the first digit
The first digit can be 1-9 [9 possibilities]
The second digit can be 0,1-9(except the first digit) [9 possibilities]
The third digit can be any of the 8 digits that are not the first and second digits [8 possibilities]
Total possible 3 digit numbers are 9*9*8 = 81*8 = 648

Total numbers(without repeated digits) between 0 and 1000 are $$9+81+648 = 738$$

Therefore, the numbers which have repeated digits are 1000 - 738 = 262(Option C)

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How many numbers from 1 to 1000, both inclusive, have digits repeated [#permalink]

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15 Apr 2018, 00:39
Hi eddwwaarrd

Please look at the solution below.

There are in total 1000 Numbers.

Now to find all the numbers that have repeated digits,we can subtract all the numbers that do not have repeated digits from the total 1000.

1-9 --> 9 numbers which are distinct
10-99--> Take this as a form of AB, where at A place you have 9 options(1-9 excluding 0) and at B place you have 9 options(0 can come up)
So total 9*9=81
100-999--> In the same manner ,Take this as a form of ABC, where at A place you have 9 options(1-9 excluding 0) and at B place you have 9 options(0 can come up) and at C place you have 8 options
So total 9*9*8=648

1000 cannot be considered since it has 0 as the digit repeating.

Total cases:- 9+81+648=738
Note 738 are those numbers that do not have repeating digits

Therefore, numbers having repeated digits are 1000-738=262
IMO C.
Hope it helps.
How many numbers from 1 to 1000, both inclusive, have digits repeated   [#permalink] 15 Apr 2018, 00:39
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