math
posted by Andrew .
if only digits 0,1,2,3,4,5,6, and 7 may be used find number of possibilites four digit numbers

n!/r!(nr)!
8!/4!(84)!
= 70 
Ok, Andrew
in my last answer to the same question I somehow missed the 7 and saw only 6 numbers.
(Have to concentrate more on reading the question more carefully)
In each case we can assume that no whole number starts with a zero, and no repeating digits for all 3 cases.
We have 8 digits, but can only use 7 in the first position ...
number of 4 digits numbers = 7x7x6x5 = 1470
number of 3 digit odd numbers, (must end in an odd) = 6x6x4 = 144
number of 3 digit numbers = 7x7x6 = 294