Math, asked by rjyuvapriya, 11 months ago

how many numbers from 1 to 2000 have none of their digits repeated?

Answers

Answered by anwar2030
0

Answer:

this question is wrong

Answered by smithasijotsl
0

Answer:

Total numbers from 1 to 2000 have none of their digits repeated = 1243

Step-by-step explanation:

To find,

The total count of numbers from 1 to 2000, with none of their digits repeated

Solution:

Number of one-digit numbers between and 1 to 2000 = 9

Number of two-digit numbers between and 1 to 2000

Number of possible digits in tens place = 9, (digit '0' is not possible)

Number of possible digits in ones place = 9, (9 digits including '0' and excluding the digit in the tens place)

Total possible two-digit numbers are = 9×9 = 81

Number of three-digit numbers between and 1 to 2000

Number of possible digits in hundred's place = 9, (digit '0' is not possible)

Number of possible digits in tens place = 9, (9 digits including '0' and excluding the digit in the hundreds place)

Number of possible digits in ones place = 8, (excluding the digits in hundreds and tens place)

The total possible three-digit numbers = 9×9×8 = 648

Number of four-digit numbers between and 1 to 2000

Number of possible digits in the thousand's place = 1(only digit 1  is possible)

Number of possible digits in hundred's place = 9, (all digits including '0' and excluding '1')

Number of possible digits in tens place = 8, (excluding the digits in hundreds and tens place)

Number of possible digits in ones place = 7, (excluding the digits in hundreds, tens place and one's place)

The total possible four-digit numbers  are 9×8×7×1  = 504

Four-digit numbers when 2000 is included = 505

Total possible numbers are 505+648+81+9 = 1243

Total numbers from 1 to 2000 have none of their digits repeated = 1243

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