How many natural numbers from 1 to 1000 have none of their digit repeated?
Answers
If you are dealing with numbers from 1 to 1000 you have:
A. 1 digit numbers
B. 2 digit numbers
C. 3 digit numbers
The one four digit number 1,000 has three repeated zeros so you can just exclude it off the bat.
The answer will be the sum of all three: A + B + C.
A:
One digit numbers by definition have zero repeats possible because there is only 1 number! Here you have 9 possible choices: 1 to 9... (the zero is not allowed because the range starts at 1).
Therefore A = 9 x 1 = 9
B:
2 digit numbers - you have 9 options for the first digit, the tens column- again you cannot choose zero because that would make the number a 1 digit number and they have been dealt with already: You can choose any number from 1 to 9. Then for every one of those you have 9 options for the second units number (0 to 9 which is 10 numbers, minus the 1 number chosen in the first number) Therefore B = 9 x 9 = 81
Note that A + B = 9 x 9 + 9 = 10x9... So you could have concatenated A and B... but I am being explicit to show the working and thought process.
C:
3 digit numbers: You have 9 options for the first digit, then 9 for the second (same argument as before then a choice of 8 digits for the third (because the third can be any number from 0 to 9 except the two chosen in the first two) = 9 x 9 x 8 = 648.
Therefore C = 9 x 9 x 8 = 648.
Total numbers is therefore: 9 + 81 + 648 = 738.
Note that when you put A + B + C together you get:
9 + 9 x 9 + 9 x 9 x 8 = 9 x 9 (1 + 8) + 9 = 9 x 9 x 9 + 9.