In how many ways can $5$ boys and $5$ girls stand in a queue such that exactly four of the girls stand consecutively in the queue? I have two combinatorics questions.

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then, the value of $m/n$ is?
Let $n_1<n_2<n_3<n_4<n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5 = 20$. The number of distinct arrangements of ($n_1, n_2, n_3, n_4, n_5$) is?

So for the first question, I was able to find out $ n = 6! × 5!$ but while finding $m$, I could do $4! × 7!$ But that also include all the fives girls accidentally coming together.. So how do I eliminate those cases? I can't proceed further.
The second one, I thought was one of those stars and bars problems, so I did $C(24, 4)$ but then I realized that the condition on the values of numbers is not that simple. I tried converting it into simpler problem, as they do with stars and bars problems but I couldn't achieve something useful. I'm stuck :/
Can you help me get through these questions?
P.S.-These questions are sometimes meant to be done using a trick, so if you think you know some trick to make that easier, please be sure to tell it. And otherwise a true solution would be as helpful :)


In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue?

You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue.  
Method 1:  We treat the block of five girls as a single object.  We then have six objects to arrange, the block of girls and the five boys.  The objects can be arranged in $6!$ ways.  The five girls can be arranged within the block in $5!$ ways.  Thus, there are $6!5!$ ways for five boys and five girls to stand in a queue if all five girls stand consecutively in the queue.
Method 2:  Line up the five boys, which can be done in $5!$ ways.  This creates six spaces in which to place the block of five girls, four between successive boys and two at the ends of the row.
$$\square b_1 \square b_2 \square b_3 \square b_4 \square b_5 \square$$
Choose one of these six spaces in which to place the block of girls, then arrange the five girls within the block.  This can be done in $6 \cdot 5!$ ways.  Hence, the number of admissible arrangements is $6!5!$.

In how many ways can $5$ boys and $5$ girls stand in a queue if exactly four girls stand consecutively in the queue?

We modify the second method above.
Line up the five boys in $5!$ ways.  This creates six spaces in which to place the girls.  Choose which four of the five girls stand consecutively, which can be done in $\binom{5}{4}$ ways.  Choose which of the six spaces the block of four girls fills.  Arrange the four girls in that space in $4!$ ways.  That leaves five spaces in which to place the remaining girl.  Hence, the number of ways five boys and five girls can stand in a queue if exactly four girls stand consecutively is 
$$5!\binom{5}{4} 6 \cdot 4! \cdot 5 = 5! \cdot 5 \cdot 6 \cdot 5 \cdot 4! = 5 \cdot 6!5!$$

In how many ways can $20$ be expressed as the sum of five distinct increasing positive integers?

Since $20$ is a small number, we can simply write down all the possibilities:
\begin{align*}
20 & = 1 + 2 + 3 + 4 + 10\\
   & = 1 + 2 + 3 + 5 + 9\\
   & = 1 + 2 + 3 + 6 + 8\\
   & = 1 + 2 + 4 + 5 + 8\\
   & = 1 + 2 + 4 + 6 + 7\\
   & = 1 + 3 + 4 + 5 + 7\\
   & = 2 + 3 + 4 + 5 + 6
\end{align*}
Notice that any sum of five distinct positive integers is at least $1 + 2 + 3 + 4 + 5 = 15$.  We then have to distribute five more ones in such a way that we preserve the increasing sequence.  Since $5$ can be partitioned into at most five positive integers in the following seven ways, 
\begin{align*}
5 & = 5\\
  & = 4 + 1\\
  & = 3 + 2\\
  & = 3 + 1 + 1\\
  & = 2 + 2 + 1\\
  & = 2 + 1 + 1 + 1\\
  & = 1 + 1 + 1 + 1 + 1
\end{align*}
we can do so in the following ways:
\begin{align*}
(0, 0, 0, 0, 5)\\
(0, 0, 0, 1, 4)\\
(0, 0, 0, 2, 3)\\
(0, 0, 1, 1, 3)\\
(0, 0, 1, 2, 2)\\
(0, 1, 1, 1, 2)\\
(1, 1, 1, 1, 1)\\
\end{align*}
Adding these, respectively, to the vector $(1, 2, 3, 4, 5)$ yields the solutions
\begin{align*}
(1, 2, 3, 4, 10)\\
(1, 2, 3, 5, 9)\\
(1, 2, 3, 6, 8)\\
(1, 2, 4, 5, 8)\\
(1, 2, 4, 6, 7)\\
(1, 3, 4, 5, 7)\\
(2, 3, 4, 5, 6)
\end{align*}
that correspond to the seven sums we wrote above.
