Confusing probability problems based on product rule and combinations I am going thru probability exercise. Faced first problem:

Book Q1. Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. What is the probability the largest number appearing on the selected tickets is 7?

My logic: if one of six tickets should be 7, the $\color{red}{\text{remaining 5}}$ can be any of 1 to 7, so it should be $7^5$. 
But turns out that the given solution is $\frac{7^6-6^6}{10^6}$. 
My Q1. Though I understood the logic behind $\frac{7^6-6^6}{10^6}$, I was wondering what is exact logical mistake with $7^5$? I guessed that $7^5$ completely ignores what should be 6th ticket, it only puts restriction on 5 tickets. Is it like that?
Then I came across similar but more involved problem, with significant difference from above one that it performs action without replacement:

Book Q2. Three numbers are chosen at random without replacement from (1,2,3,...,10). What is the probability that the minimum number is 3 or the maximum number is 7?

My logic: Noticing that this is without replacement, I guessed the solution should be 
$$
=
\begin{pmatrix} 
\text{selections with}\\ 
\text{minimum}\\ 
\text{number is 3} 
\end{pmatrix}
+
\begin{pmatrix} 
\text{selections with}\\ 
\text{maximum}\\ 
\text{number is 7} 
\end{pmatrix} 
-
\begin{pmatrix} 
\text{selections with}\\ 
\text{maximum}\\ 
\text{number is 7}\\ 
\text{and minimum}\\ 
\text{number is 3} 
\end{pmatrix} 
$$
$$=
\frac{
\overbrace{(\binom{8}{3}-\binom{7}{3})}^{\text{#selections with min 3}}
+
\overbrace{(\binom{7}{3}-\binom{6}{3})}^{\text{#selections with max 7}}
-
\overbrace{3\times {^3P_3}}^{\text{#selections with max 7 and min 3}}
}
{\binom{10}{3}
}
$$
But the book solutions says:

P(minimum 3) or P(maximum 7)
P(minimum 3) $=\frac{\binom{7}{2}}{\binom{10}{3}}=\frac{21}{120}$
P(max 7) $=\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{120}$
Thus the solution is $\frac{11}{40}$

My Q2. How even by books logic the solution $\frac{11}{40}$ is achieved. I am unable to understand it as I find the explanation insufficient.
My Q3. If book Q2 answer is correct then why for book Q1 solution is not $7^5$ which is what I initially guessed (because the only difference being with / without replacement, the logic of getting $\color{red}{\text{remaining m}}$ stuffs out n should remain same)?
My Q4. If we make first question without replacement, will the solution be $\frac{\binom{7}{6}-\binom{6}{6}}{\binom{10}{6}}$?
My Q5. What will be the solution if we make book Q2 with replacement?
My Q6. Where my logic for solution to Book's Q2 is wrong?
 A: Addressing your Q1, the difference between the answer you gave and the answer that the book gives is that your solution assumes that a specific selection is 7.  That is, $\frac{7^5}{10^6}$ is the probability that a fixed ticket is 7 and the other five take values from 1 to 7 while the books solution just assumes that at least one of the tickets is 7.
Q2 The second problem is much different as there is no order on the sets of chosen numbers.  Note that $\binom{8}{3}-\binom{7}{3}=\binom{7}{2}$  and $\binom{7}{3}-\binom{6}{3}=\binom{7}{2}$, so that part of your answer is consistent with the book solution--though the logic that the book uses might be a tad bit more straightforward.  In each of the above cases, the book solution fixes the given number and then chooses two other numbers at random from the allowed set (either $\{$greater than 3$\}$ or $\{$less than 7$\}$).  Using this type of reasoning we can quickly enumerate the number of choices with max 7 and min 3: two of the choices are fixed and there are only three possibilities for the third choice.  Hence the solution is
$$
\frac{\binom{7}{2}+\binom{6}{2}-3}{\binom{10}{3}}=\frac{11}{40}.
$$
For your Q3, again, the implied order in the first problem makes it easy to accidentally specify an order during the solution process (and as a result get the wrong answer).  Note that your solution for the second problem was not necessarily wrong (I don't know what $3\times^3P_3$ means), you just used a different method than the book.
For your Q4 you are correct.  This is the same thing as $$\frac{\binom{7}{5}}{\binom{10}{6}}.$$
Q5  The answer would be
$$
\frac{(8^3-7^3)+(7^3-6^3)-5}{10^3}.
$$
Let me know if you have any questions about how I arrived at the above answer.
A: With the actual answers already given, I am putting different possible scenarios and there solutions here for reference.


*

*Selecting six out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 = ?
Sol. 
$$
\overbrace
{\frac{7^6-6^6}{10^6}}
^{
\begin{pmatrix}
\text{no. of ways of selecting any six out of {1,..,7}} \\
\text{removing those not containing 7}\\
\text{i.e. those six from {1,..,6}}\\
\end{pmatrix}
}=0.070993$$

*Selecting six out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 = ?
Sol. 
$$
\frac{
\overbrace{\binom{7}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,7}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
-
\overbrace{\binom{6}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,6}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
}{
\binom{10}{6}
}=
\frac{
\overbrace{\binom{6}{5}}
^{\begin{pmatrix}
\text{selecting six from {1,..,7}} \\
\text{with 7 as max reduces to} \\
\text{selecting 5 from {1,..,6}} \\
\text{since you know you must} \\
\text{have 7} \\
\end{pmatrix}}
}{\binom{10}{6}}=\frac{1}{35}=0.028571
$$

*Selecting six out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 = ?
Sol.
$$
\frac
{\overbrace
{^6P_5}
^{
\begin{pmatrix}
\text{no. of ways to select five} \\
\text{out of {1,..,6} while} \\
\text{considering order} \\
\end{pmatrix}
}
\times
\overbrace
{6}
^{
\begin{pmatrix}
\text{putting 7} \\
\text{in one of 6 places} \\
\end{pmatrix}
}
}{^{10}P_6}
$$

*Selecting three out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$\frac{(8^3-7^3)=(7^3-6^3)-18}{10^3}=0.278$$
For details, see correction in Brian Scott's solution.

*Selecting three out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{^7P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {4,..,10}}  and \\
\text{put 3 in any of three places} \\
\text{making 3 min number among} \\
\text{three selected} \\
\end{pmatrix}
}
+
\overbrace
{^6P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {1,..,6} put 7}  and \\
\text{put 7 in any of three places} \\
\text{making 7 max number among} \\
\text{three selected} \\
\end{pmatrix}
}
-
\overbrace
{^3P_3 \times 3}
^{
\begin{pmatrix}
\text{Selecting 3 tickets with both} \\
\text{min 3 and max 7:} \\
\text{first select any one of} \\
\text{4,5 or 6,then perform} \\
^3P_3\text{ arrangements with the} \\
\text{selected number,3 & 7} \\
\end{pmatrix}
}
}{^{10}P_3}
= \frac{198}{720} = 0.275
$$

*Selecting three out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{\binom{8}{3}-\binom{7}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 3 as minimum} \\
\end{pmatrix}
}
+
\overbrace
{\binom{7}{3}-\binom{6}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 7 as maximum} \\
\end{pmatrix}
}
-
\overbrace
{3}
^{
\begin{pmatrix}
\text{no. of selections with both} \\
\text{3 as min & 7 as max:} \\
\text{{3,7,4},{3,7,5},{3,7,6}} \\
\end{pmatrix}
}
}{\binom{10}{3}}
=\frac{\binom{7}{3}+\binom{6}{2}-3}{\binom{6}{2}}
=\frac{11}{40}
=0.275
$$
