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

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

I don't get this. Can someone explain more? Somebody please. Thanks


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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.

Total-1000 numbers

Calculate numbers in which no digit is repeated

1 digit number-0 to 9= 9 cases
2 digit cases -9*9=81 cases

As 0 can't come in tens digit and digit in tens digit can't come at units place

3 digit cases-9*9*8=648

1000-(9+81+648)=262

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