Am I on the right track with these questions?
Find expressions for each of the following. (Leave your answer as a mathematical expression rather than a number.)
1) The number of strings of $7$ lower case letters (a–z) that do not contain any letter twice or more.
I believe 1 is an ordered selection without repetition that is represented by $$\frac{26!}{(26-7)!}$$
2) The number of binary strings of length $50$ that contain at most three 1s.
I believe 2 is an unordered selection with repetition that is represented by
$$\frac{(50 + 3 - 1)!}{3!(50 - 1)!} + \frac{(50 + 2 - 1)!}{2!(50 - 1)!} + \frac{(50 + 1 - 1)!)}{1!(50 - 1)!}$$
3) The number of ternary strings of length $10$ (strings of 0s, 1s and 2s) containing exactly one 1 and exactly four 2s.
I believe this is ordered with repetition but I am not sure how I would represent this.
Is my way of thinking correct?
Your interpretation that this is a permutation of $7$ distinct letters selected from the $26$ lower case letters of the alphabet is correct, as is your answer.
There are four cases. The binary string contains no 1s, one 1, two 1s, or three 1s. The number of binary strings of length $50$ that contain exactly $k$ 1s is $\binom{50}{k}$. Hence, the number of binary strings of length $50$ that contain at most three 1s is $$\sum_{k = 0}^{3} \binom{50}{k} = \binom{50}{0} + \binom{50}{1} + \binom{50}{2} + \binom{50}{3}$$
Notice that a ternary string of length $10$ that contains exactly one 1 and exactly four 2s must contain exactly $10 - 1 - 4 = 5$ 0s. Choose one of the ten positions for the 1, four of the remaining nine positions for the four 2s, and all five of the remaining five positions for the five 0s. $$\binom{10}{1}\binom{9}{4}\binom{5}{5}$$ This expression can be simplified as follows: $$\binom{10}{1}\binom{9}{4}\binom{5}{5} = \frac{10!}{1!9!} \cdot \frac{9!}{4!5!} \cdot \frac{5!}{5!0!} = \frac{10!}{1!4!5!0!} = \frac{10!}{1!4!5!}$$ since $0! = 1$. While $1! = 1$, I chose to leave it in the denominator so that you could see that the terms in the denominator correspond to the number of appearances of each digit in the ternary string. The factor $4!$ represents the number of ways the four 2s can be permuted among themselves within an arrangement without producing an arrangement distinguishable from the original arrangement. The factor $5!$ represents the number of ways the five 0s can be permuted among themselves within an arrangement without producing an arrangement distinguishable from the original arrangement.