d) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?(

The question is incomplete. Here is the complete question.

(a) How many three-digit numbers can be formed from the digits 0,1,2,3,4,5 and 6, if each digit can be used only once?

(b) How many of these are odd numbers?

(c) How many are greater than 330?

(a) 180

              (b) 75

               (c) 105

Step-by-step explanation:

(a) In the group, there are 7 digits. A three-digit number can not start with zero, otherwise, it will be a 2-digit number. So:

For the hundreds position, there are 6 choices.

For the tens position, since the digit can be used only once, there are 6 choices.

For the unit position, there are 5 choices.

The total three-digit number formed is: 6*6*5 = 180

(b) To form an odd number, the unit position must be an odd digit, then:

unit position has 3 choices;

hundreds position has 5 choices;

tens position has 5 remaining choices.

The total three-digit odd number is: 3*5*5 = 75

(c) The number formed must be greater than 330, so:

If the number start with a 3, to be greater, there are 3 other choices (4, 5 and 6), so Tens position has 3 choices and Unit position has 5 choices.

Total number is: 3*5 = 15

Another possibility is the number starts with a digit bigger than 3 and so,  there are 3 choices.

Tens position has 6 choices;

Unit position has 5 choices;

Total possibilities are: 3*6*5 = 90

The total number of ways a three-digit number is greater than 330 is:

90 + 15 = 105

don't know sorry bru

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)