Permutation question Leading zeros are not allowed
1) How many seven-digit numbers have no repeated digits? Would this one be $9*P(9,6)$
2) How many seven-digit numbers with no repeated digits contain a 3 but not a 6? And would this be  $8*P(9,6)$
 A: For 1, you are correct.  The first digit has 9 choices, then the rest have $P(9,6)$ to deal with the no duplicates requirement
For 2, you have done nothing to require a 3, nor to prohibit a 6 except in the first digit.
A: Some hints for the first problem:
How many ways can you fill the millions place?  (The answer is $9$.  Why?)  Then, how many ways can you fill the hundred thousands place (without repeating anything)?  (Again, it's $9$.  Why?)  What about the ten thousands place?  (This time it's $8$.)  And so forth.  Multiply them together.
For the second problem:
Since you can't have leading zeroes, try two cases.  You know you need a $3$ in there, so make that the millions place.  Then how many ways can you build up that number without repeating digits?  Now, make the first digit something other than $3$ (but not $0$ or $6$, of course).  You'll still need a $3$ later in the number.  How many ways can you build up this one?  Add these two totals together.
A: question 1: 
$9*9*8*7*6*5*4$
question 2:
case 1: number 3 is in first digit: then $8*7*6*5*4*3$
case 2: number 3 is not in first digit: then
$6*7*7*6*5*4*3$
A: 1) There are $9$ choices for the first digit. For every choice of first digit, there are $9$ allowed choices for the second digit, since $0$ is now allowed, the only thing forbidden is the number we used for the first digit.
For every choice of first two digits, there are $8$ allowed choices for the third digit. And so on.
2) Here things are a little different if the first digit is a $3$, or not a $3$.
If the first digit is a $3$, that leaves $8$ allowed digits out of which to make the six-digit tail end (recall $6$ is not allowed). There are $(8)(7)(6)(5)(4)(3)$ such tail ends. 
If the first digit is not a $3$, there are $7$ allowed choices.  Now we need to place a $3$ somewhere. Where can be chosen in $6$ ways. That leaves $5$ empty slots, which can be filled from allowed digits in $(7)(6)(5)(4)(3)$ ways.  
Now we put the pieces together. 
Another way: For 2), we can easily imitate the process of 1) to get the number of strings that miss $6$. Call this number $x$.
Now count the strings that miss both $6$ and $3$. Call this number $y$.
The answer to 2) is then $x-y$. 
