Using only the digits 0 an 1, how many different numbers consisting of 8 digits can be formed?
Can your leading digit be 0 ?
if so, then the number of cases = 2^8 = 256
if not, the the number of cases = 1*2^7 = 128
(looks like you are studying the binary number system)
yeah its the first one. thank you!
To determine how many different numbers consisting of 8 digits can be formed using only the digits 0 and 1, we can use the concept of permutations.
Since each digit can only be 0 or 1, and there are 8 positions to fill, we have 2 choices for each position. Therefore, the total number of different numbers can be calculated using the formula:
Total number of different numbers = 2^8
Using exponents, we can simplify this equation:
2^8 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256
Therefore, there are 256 different numbers consisting of 8 digits that can be formed using only the digits 0 and 1.