Some more questions, I believe I have them right but I have no way to check my answers.
The way I think this one is answered is by filling in the 10 spots that have to be 7 first. That gives you 45 spots for 8 possible numbers... so the answer is simply $8^{45}$?
How many ways are there to pick a dozen donuts from 5 types?
$5^{12}$
(Sorry I'm on my phone or I would use latex or whatever)
On
Suppose all the 7s are the 10 firsts numbers of our string. So we still have 45 "bits" that we can fill with numbers in the set $\{1,..,6,8,9\}$ which has 8 elements. So, we can put each number of this set in each bit of our string. This gives to us $8^{45}$ possible strings. BUT, remember we fixed the positions of the 7s. How many positions we can chose to the ten 7s? Well, $ 55 \choose 10$. So the answer should be ${55}\choose{10}$$8^{45}$.
Now, for the second question we can translate it in another language. Let $x_i$ be the number of donuts of type $i$ we have chose. So, the quantities $x_i$'s must satisfy $$ x_1 + ... + x_5 = 12$$
But, you can make your choice in many ways. You can decide to don't to chose the donut of type 1, in this case you have $x_1 = 0$
This link http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29 can help you how to get the answer $ {5+12-1} \choose {12-1}$
$$\binom {55}{10}9^{55-10}\text{ and }\binom{5+12-1}{12-1}$$ Choose 10 positions for sevens, fill them, choose the remaining numbers, for the last use stars and bars.