a 9-character numeric passcode is constructed at random. if the last number cannot be prime and the first digit cannot be 0, 1, or 7, how many different codes can be constructed?

The members of the set "0, 1" make up the digits in the binary number system. Computers employ this technique because the two numbers can stand in for the low and high logic states. In computer jargon, "binary digit" is abbreviated to "bit.The repeated digits 0 1 2 3 can be used to generate how many four digits?The third position therefore has two possibilities and can be filled by any of the two digits. One digit will be left once the third blank has been filled. As a result, there is only one way to fill in the fourth blank, which is the last digit. Thus, there are 18 four-digit integers that may be created using the digits 0, 1, 2, and 3.For example  : The initial digit has seven possible values because [0, 1 and 7] are not included.There are six potential values because the last digit excludes [2, 3, 5 and 7].Seven of the first eight numbers can include any of the 10 values.Therefore different codes that can be constructed is 7*10*10*10*10*10*10*10*6 = 7*6*(10^7) = 420,000,000So, 420,000,000 is the correct response.To Learn more About binary digit refer to :https://brainly.com/question/20855434#SPJ4