Discrete Math - Combinatorics related question - with repetitions 
A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose 23 croissants with at least five chocolate croissants and at least three almond croissants?

If we choose $23$ croissants and each time we have selected $5$ chocolate and $3$ almond, we are left with $23 - 8 = 15$ choices.
So we have $4$ remaining croissants, $n = 4, r = 15$ and using the formula :  $\binom{n+r-1}{r}$
We have $\binom{4+15-1}{15}$
$\binom{18}{15} = 816$
We have $816$ different ways to choose $23$ croissants which should have at least $5$ chocolate and $3$ almond.
I am thinking since it does not specify a higher limit, it could have more but does the calculation I have shown suffice the requirement? Please advise.
 A: 
A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants.  How many ways are there to choose $23$ croissants with at least five chocolate croissants and at least three almond croissants?

The problem asks us to find the number of ways to select $23$ croissants from the six flavors offered by the shop if at least five chocolate croissants and at least three almond croissants are selected.  That means more than five chocolate or more than three almond croissants may be selected.  Consequently, we are allowed to select the $23 - 5 - 3 = 15$ additional croissants from all six flavors the shop offers after selecting five chocolate and three almond croissants.
Let $x_i$, $1 \leq i \leq 6$, be the number of additional croissants of flavor $i$ we select.  Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 15 \tag{1}$$
is an equation in the nonnegative integers.  A particular solution of equation 1 corresponds to the placement of $6 - 1 = 5$ addition signs in a row of $15$ ones.  For instance,
$$1 1 + 1 1 1 + 1 1 1 1 1 + + 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 2, x_2 = 3, x_3 = 5, x_4 = 0, x_5 = 3, x_6 = 2$.  The number of solutions of equation 1 is the number of ways we can place $6 - 1 = 5$ addition signs in a row of $15$ ones, which is
$$\binom{15 + 6 - 1}{6 - 1} = \binom{20}{5}$$
since we must choose which five of the $20$ positions required for $15$ ones and $5$ addition signs will be filled with addition signs.
The number you found is indeed a lower bound since you only considered the case in which exactly five chocolate and exactly three almond croissants are included in the selection of $23$ croissants.
