Number of ways to pick 12 elements from 3 different kinds of different no of elements Question:
A florist has 5 aspidistras, 6 buttercups, and 7 chrysanthemums. How many different kinds of bouquets of a dozen flowers (it is not required to use all types of flowers) can she make from these?
My Solution:
The solution I came up with is as follows:
Since it seems that the order of flowers doesn't matter, thus we can just add them all up and choose a dozen flowers out of them.
5 aspidistras + 6 buttercups + 7 chrysanthemums = 18 total flowers
answer = 18 choose 12 = 18564
Is the above answer correct or if not can anyone tell me what I am doing wrong? Any help is greatly appreciated.
 A: I count 27 different bouquets.
Let A,B,C denote the different flower names.
Ways that include only two types of flowers:


*

*--- 7 C's and 5 A's

*--- 7 C's and 5 B's

*--- 6 C's and 6 B's


Ways that include three types of flowers:


*--- 7 C's, 4 A's, 1 B

*--- 7 C's, 4 B's, 1 A

*--- 7 C's, 3 A's, 2 B's

*--- 7 C's, 3 B's, 2 A's

*--- 6 C's, 5 A's, 1 B

*--- 6 C's, 5 B's, 1 A

*--- 6 C's, 4 A's, 2 B's

*--- 6 C's, 4 B's, 2 A's

*--- 6 C's, 3 A's, 3 B's

*--- 6 B's, 5 A's, 1 C

*--- 6 B's, 5 C's, 1 A

*--- 6 B's, 4 A's, 2 C's

*--- 6 B's, 4 C's, 2 A's

*--- 6 B's, 3 A's, 3 C's

*--- 5 C's, 5 B's, 2 A's

*--- 5 C's, 5 A's, 2 B's

*--- 5 B's, 5 A's, 2 C's

*--- 5 C's, 4 B's, 3 A's

*--- 5 C's, 4 A's, 3 B's

*--- 5 B's, 4 A's, 3 C's

*--- 5 B's, 4 C's, 3 A's

*--- 5 A's, 4 B's, 3 C's

*--- 5 A's, 4 C's, 3 B's

*--- 4 A's, 4 B's, 4 C's

A: Let $a$ be the number of aspidistras, $b$ be the number of buttercups, and $c$ be the number of chrysanthemums.  The number of ways of forming a bouquet of a dozen flowers is the number of solutions of the equation
$$a + b + c = 12 \tag{1}$$
in the nonnegative integers subject to the constraints $a \leq 5$, $b \leq 6$, and $c \leq 7$.
A particular solution of equation 1 corresponds to the placement of two addition signs in a row of twelve ones.  For instance,
$$1 1 1 1 + 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $a = 4$, $b = 5$, and $c = 3$, while 
$$+ 1 1 1 1 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $a = 0$, $b = c = 6$.  The number of such solutions is 
$$\binom{12 + 2}{2} = \binom{14}{2}$$
since we must choose which two of the fourteen positions required for twelve ones and two addition signs will be filled by addition signs.
In general, the equation 
$$x_1 + x_2 + \ldots + x_k = n$$
has 
$$\binom{n + k - 1}{k - 1}$$
solutions in the nonnegative integers since a particular solution corresponds to the placement of $k - 1$ addition signs in a row of $n$ ones, and we must choose which $k - 1$ of the $n + k - 1$ positions required for $n$ ones and $k - 1$ addition signs will be filled with addition signs.
Now, we must subtract the number of solutions that violate one or more of the constraints.  Notice that it is not possible to violate two constraints simultaneously since $6 + 7 = 13 > 12$.
Suppose $a > 5$.  Then $a' = a - 6$ is a nonnegative integer.  Substituting $a' + 6$ for $a$ in equation 1 yields 
\begin{align*}
a' + 6 + b + c & = 12\\
a' + b + c & = 6 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{6 + 2}{2} = \binom{8}{2}$$
solutions.   
Suppose $b > 6$.  Then $b' = b - 7$ is a nonnegative integer.  Substituting $b' + 7$ for $b$ in equation 1 yields 
\begin{align*}
a + b' + 7 + c & = 12\\
a + b' + c & = 5 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with 
$$\binom{5 + 2}{2} = \binom{7}{2}$$
solutions.
Suppose $c > 7$.  Then $c' = c - 8$ is a nonnegative integer.  Substituting $c' + 8$ for $c$ in equation 1 yields 
\begin{align*}
a + b + c' + 8 & = 12\\
a + b + c' & = 4 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with 
$$\binom{4 + 2}{2} = \binom{6}{2}$$
solutions.  
Hence, the number of bouquets that can be formed is 
$$\binom{14}{2} - \binom{8}{2} - \binom{7}{2} - \binom{6}{2} = 27$$
