First put $w$ balls in each of the $n$ boxes; that uses up $wn$ of the balls, leaving you $r-wn$ to distribute arbitrarily amongst the $n$ boxes. By the standard stars-and-bars argument, the number of ways to distribute $m$ indistinguishable balls amongst $n$ distinguishable boxes is
$$\binom{m+n-1}{n-1}\;,\tag{1}$$
and in this case $m=r-wn$, so in this case we get
$$\binom{r-wn+n-1}{n-1}\;.$$
(Note that you have a sign wrong in the expression at the beginning of your question.)
Now recall the identity $\binom{n}k=\binom{n}{n-k}$ and apply it to $(1)$ to get
$$\binom{m+n-1}{n-1}=\binom{m+n-1}m\;.\tag{2}$$
Thus, the righthand side of $(2)$ also gives the number of ways to distribute $m$ indistinguishable balls amongst $n$ boxes (and you have another sign error: your $n-k+1$ should be $n+k+1$). In the case $m=r-wn$ in which we start with the minimum of $w$ balls in each box, it becomes
$$\binom{r-wn+n-1}{r-wn}\;.$$
In the specific problem at hand, however, you can’t use either of these without some adjustment. You have $6$ boxes, namely, the $4$ slots between adjacent vowels and the slots on the two ends, and $21$ balls. Each of the $4$ boxes in the middle must receive at least one ball, but either or both of the end slots could be left empty. Thus, you should begin by putting one ball in each of the $4$ middle boxes, leaving $17$ balls to be distributed arbitrarily amongst the $6$ boxes. Either expression in $(2)$ can now be used: there are $$\binom{17+6-1}{6-1}=\binom{22}5=\binom{22}{17}=\binom{17+6-1}{17}$$ ways to distribute the consonants amongst the vowels, ignoring which consonant is which. Of course this must then be multiplied by $5!21!$ to account for the order of the vowels and consonants within their respective sets of positions in the string.
First problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the number of balls in each box is odd?
Let $x_k$ denote the number of balls placed in the $k$th box, $1 \leq k \leq 6$. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{1}$$
Since each $x_k$ is a positive odd integer, $x_k = 2y_k + 1$ for some non-negative integer $y_k$. Substituting into equation 1 and simplifying yields
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 10 \tag{2}$$
which is an equation in the non-negative integers. The number of solutions of equation 2 is the number of ways five addition signs can be inserted into a row of ten ones.
There are $$\binom{10 + 5}{5} = \binom{15}{5}$$ solutions since we must choose which five of the fifteen symbols (ten ones and five addition signs) will be addition signs.
Second problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the first three boxes contain at most six balls?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{3}$$
in the non-negative integers subject to the constraints that $x_1, x_2, x_3 \leq 6$. Equation 3 is an equation in the non-negative integers. It has
$$\binom{26 + 5}{5} = \binom{31}{5}$$
solutions.
We must exclude those cases in which one or more of the constraints is violated.
Suppose the constraint $x_1 \leq 6$ is violated. Then $x_1 \geq 7$. Let $y_1 = x_1 - 7$. Then $y_1$ is a non-negative integer. Substituting $y_1 + 7$ for $x_1$ in equation 3 and simplifying yields
$$y_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 19 \tag{4}$$
Equation 4 is an equation in the non-negative integers. It has
$$\binom{19 + 5}{5} = \binom{24}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if the constraint $x_2 \leq 6$ or the constraint $x_3 \leq 6$ is violated. Hence, there are
$$\binom{3}{1}\binom{24}{5}$$
solutions in which one of the constraints is violated.
Suppose the constraints $x_1 \leq 6$ and $x_2 \leq 6$ are both violated. Let $y_1 = x_1 - 7$; let $y_2 = x_2 - 7$. Then $y_1$ and $y_2$ are non-negative integers. Substituting $y_1 + 7$ for $x_1$ and $y_2 + 7$ for $x_2$ and simplifying yields
$$y_1 + y_2 + x_3 + x_4 + x_5 + x_6 = 12 \tag{5}$$
which is an equation in the non-negative integers with
$$\binom{12 + 5}{5} = \binom{17}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if both the constraints $x_1 \leq 6$ and $x_3 \leq 6$ are violated or both the constraints $x_2 \leq 6$ or $x_3 \leq 6$ are violated. Hence, there are
$$\binom{3}{2}\binom{17}{5}$$
solutions in which two of the constraints are violated.
Suppose that all three constraints are violated. Let $y_k = x_k - 7$, $1 \leq k \leq 3$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 7$ for $x_k$, $1 \leq k \leq 3$, and simplifying yields
$$y_1 + y_2 + y_3 + x_4 + x_5 + x_6 = 5 \tag{6}$$
which is an equation in the non-negative integers with
$$\binom{5 + 5}{5} = \binom{10}{5}$$
solutions.
By the Inclusion-Exclusion Principle, the number of ways the $26$ identical balls can be distributed into six distinct boxes such that the first three boxes contain at most six balls is
$$\binom{31}{5} - \binom{3}{1}\binom{24}{5} + \binom{3}{2}\binom{17}{5} - \binom{3}{3}\binom{10}{5}$$
Best Answer
For the first question:
First place the consonants in the following way $$-P-S-N-S-,$$ so, this can be done in $$4!/2!=12$$ ways. Now place the vowels in the gaps, so this can be done in $$\frac{\binom 53 \times 3!}2=30$$ ways. So, the total number of way is $$\bbox[border:2px solid red] {12\times 30=360}$$ ways.
Note: here is similar answer.
For the second one:
The order of the fishes is irrelevant, so, answer is $$\bbox[border:2px solid red] {\binom {65}{13}\times\binom {52}{13}\times\binom {39}{13}\times\binom {23}{13}\times\binom {13}{13}\\=\frac{65!}{(13!)^5}}$$
For the third one:
Here I assume that the boxes are distinct. So, you have only to count the number of blue balls in one box.
Let, $b_1,b_2,b_3,b_4,b_5$ be the three box, and define $x_i$ as the number of balls in the $i^{th}$ box. So, you have to count the number of solutions to the equation $$\sum_{i=1}^5 x_i=13,$$ which is $$\binom{17}{4}.$$ And now choose any $2$ box out of $5$ to keep the $2$ red balls, in $\binom 52$ ways, so, number of ways is $$\bbox[border:2px solid red] {\binom{17}{4}\times \binom 52}$$