How many ways are there to give $13$ identical candies to $4$ boys and $4$ girls with restrictions? How many ways are there to give $13$ identical candies to $4$ boys and $4$ girls so that:
a) No boy misses out? For this question, I thought that I would first give $1$ candy to each boy, which leaves me with $9$ candies. Then I use the formula of combination with repetition which gives me $\binom{9+8-1}{8} = \binom{16}{8} $. However the answer given to me is $\binom{16}{7} $ and I don't know why. 
b) No one misses out and no girl gets
more candies than any boy? I was given $60$ as an answer. I know that I first have to give a candy to every person, so that leaves me with $5$ candies but I don't know how to proceed from there. 
 A: For your second question, the number $13$ is very significant. You must allocate $2$ candies to each boy before you can give a second candy to any girl, so there are only $4 \choose 1$ ways in which a girl gets any candy. Therefore, you should just consider how to allocate the remaining $5$ candies to the $4$ boys to get the rest of the possibilities - and you already know how to model this problem, namely, the same way you do problem $(1)$.
These problems are both about the classical "balls in urns" formula. In my comment above, I linked you to a proof I gave of this formula elsewhere.
A: 
How many ways are there to give $13$ identical candies to $4$ boys and $4$ girls so that no boys misses out?

Since you have already given each of the four boys a candy, you have $9$ candies left to distribute to eight children.  Let $b_k$, $1 \leq k \leq 4$, be the number of additional candies received by the $k$th boy; let $g_k$, $1 \leq k \leq 4$, be the number of candies received by the $k$th girl.  Then we need to determine the number of solutions of the equation 
$$b_1 + b_2 + b_3 + b_4 + g_1 + g_2 + g_3 + g_4 = 9 \tag{1}$$
in the nonnegative integers.  A particular solution to equation 1 corresponds to the placement of seven addition signs in a row of nine ones.  For instance,
$$+ + + 1 + 1 1 + 1 1 + 1 1 + 1 1$$
corresponds to the solution $b_1 = b_2 = b_3 = 0$, $b_4 = 1$, $g_1 = g_2 = g_3 = g_4 = 2$ (each girl and the fourth boy receive two candies each, while the other boys receive one).  The number of solutions of equation 1 in the nonnegative integers is the number of ways we can place seven addition signs in a row of nine ones, which is 
$$\binom{9 + 7}{7} = \binom{16}{7}$$
since we must choose which seven of the $16$ positions (nine ones and seven addition signs) will be filled with addition signs.
The formula you used for the number of solutions of the equation 
$$x_1 + x_2 + x_3 + \cdots + x_k = n \tag{2}$$
in the nonnegative integers was incorrect.  Since a particular solution of equation 2 corresponds to the placement of $k - 1$ addition signs in a row of $n$ ones, the number of such solutions is 
$$\binom{n + k - 1}{k - 1}$$
since we must choose which $k - 1$ of the $n - k + 1$ positions ($n$ ones and $k - 1$ addition signs) will be filled with addition signs. 

How many ways are there to give $13$ identical candies to $4$ boys and $4$ girls so that no one misses out and no girl receives more candies than any boy?

Since you first gave each of the eight children a candy, you have $5$ candies left to distribute.  
It is not possible to give two or more of them to a girl since a boy would be left with fewer candies than that girl.
A girl receives an additional piece of candy:  If one additional piece of candy is given to a girl, then each boy must receive one.  The lucky girl can be selected in four ways.  
No girl receives and additional piece of candy:  If no additional pieces of candy are given to a girl, then we have five pieces of candy to be distributed to four boys.  The number of ways this can be done is the number of solutions of the equation 
$$b_1 + b_2 + b_3 + b_4 = 5$$
in the nonnegative integers.  You can use the formula stated above to determine that amount.

 $$\binom{5 + 4 - 1}{4 - 1} = \binom{8}{3}$$

Since the two cases are disjoint, the number of ways to distribute the candy can be found by adding the number of ways the candy can be distributed if a girl receives an additional piece of candy to the number of ways the candy if no girl receives an additional piece of candy.

  $$\binom{4}{1} + \binom{8}{3}$$

