We put our cryptographers to work on this apparently coded question, and asked them
to decipher what it's trying to ask. They struggled and debated among themselves for
quite some time. Then, just before lunch, they agreed that there is an 83% probability
that the question is:
"How many 3-digit numbers can be made with the common decimal digits zero through 9 ?"
The largest one is 999, so there would be 999 of them, but the first 99 don't count, because
the numbers 1 - 99 have less than 3 digits.
So there are (999 - 99) = 900 different 3-digit numbers.
1000
1000
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
you can image the 3 digits as a counter 0 0 1 0 0 2 ... ... ... 9 9 8 9 9 9 every digit uses 100 times a 6..... 100*3 digits = 300 sixes :)
648
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
1000
1000
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
9.
you can image the 3 digits as a counter 0 0 1 0 0 2 ... ... ... 9 9 8 9 9 9 every digit uses 100 times a 6..... 100*3 digits = 300 sixes :)
648
10 digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
1000
ABC A = 1,2,3,4,5,6,7,8,9; 9 possibilities B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities So there are 9*10*10 = 900 numbers with 3 digits.
Ten digits. They are: 0 1 2 3 4 5 6 7 8 9
9