A) This problem involves 8-digit binary strings such as 10011011 or 00001010.

i How many such strings are there?

ii. How many such strings end in 0?

iii. How many such strings have l's for their second and fourth digits?

iv. How many such strings have l's for their second or fourth digits?

b) Find how many 9-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6,

7, 8, 9 if repetition is not allowed and all the odd digits occur first (on the

left) followed by all the even digits.

c) How many permutations of the letters A, B, C, D, E, F, G are there in which

the three letters ABC appear consecutively, in alphabetical order?

a)

i) With an eight digit binary number, there are 2⁸, or 256 possible combinations.

ii) Each digit is set to 1 or 0 for exactly half of all combinations.  This means that half of all strings, or 256 / 2 = 128 strings end with a zero.

iii) 3/4, or 128 + 64 = 192 of the combinations have a 1 for either the second or fourth digit.  If you consider:

Half of all digits will have the second bit set to 1

Half of all digits will have the fourth bit set to 1

Half of those will overlap

Then that's half the digits, plus half the digits, minus half of half the digits. 1/2 + 1/2 - 1/4 = 3/4

iv) This is a duplicate of iii, so again, 192

b) If we're allowed to repeat the usage of the digits, then answer would be the number of values available to the power of the number of digits used.  We're allowed six distinct digits, so the total combination of 9 digit numbers would be 6⁹, or 10077696

c) In this case we can look at it simply as permutations of the set {"ABC", "D", "E", "F"}, which gives us 4! which equals 24

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

exactly you go it

step-by-step explanation: