How many numbers lying between 10 and 1000 can be formed from the
digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition is allowed)
How many numbers lying between 10 and 1000 can be formed from the
digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition is allowed)
The correct option is B 810
The total number between 10 and 1000 are 989 but we have to form the numbers by using numerals 1,2,........9, i.e., 0 is not occuring so the numbers containing any '0' would be excluded i.e., Required number of ways
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⎪⎩20,30,40,...........................100=9101,102,...........................200=19201,...............................300=19..................................................................................901,..............................900=18⎫⎪
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= 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 9×9 = 81
and number of 3 digit numbers 9×9×9 = 729
Hence total ways = 81 + 729 = 810.
810
The total number between 10 and 1000 are 989 but we have to form the numbers by using numerals 1,2,........9, i.e., 0 is not occuring so the numbers containing any '0' would be excluded i.e., Required number of ways
= 989 - ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩20,30,40,...........................100=9101,102,...........................200=19201,...............................300=19..................................................................................901,..............................900=18⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
= 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 9×9 = 81
and number of 3 digit numbers 9×9×9 = 729
Hence total ways = 81 + 729 = 810.
