How many ways are there to arrange a basket with 12 fruits comprised of 4 different kind of fruits with no more than 4 of the same kind 
In how many ways is it possible to make a basket with 12 fruits comprised of passionflower, lychee, mango and berries where the number of each kind of fruit isn't higher than 4 ?

So this is basically like having 4 different color of balls in a bin with no more then 4 balls of the same color. 
I think we need to solve this: $x_1+x_2+x_3+x_4=12$ where each $x_i\le4$ using inclusion exclusion principle and stars and bars: 
$\displaystyle{15 \choose 3}-\bigcup_{i=1}^4A_i$ where each $A_i$ is a different color. 
But now I'm not sure on how to continue.
 A: Put $4$ of each kind of fruit in the basket, which gives you $16$ pieces of fruit in all, and now ask yourself how many different ways can you take $4$ pieces out to get the number in the basket down to $12$.  This is just a stars-and-bars problem, and the answer is ${4+3\choose3}={7\choose3}=35$.
Note:  This approach works fine if the $12$ is replaced by $13$, $14$, $15$, or even $16$, in which cases you're asking how many different ways you can remove $3$, $2$, $1$, or $0$ pieces, respectively.  (The answers are $20$, $10$, $4$, and $1$.)  However, it does not work if you change the $12$ to $11$ or lower.  (The answer for $11$ is $52$, which is not a stars-and-bars type number.)
A: Personnally, I would just list all the possibilities, then for each one, count every possible permutation.
Here, you have only 5 possibilities (unless I missed one) :


*

*4 4 4 0

*4 4 3 1

*4 4 2 2

*4 3 3 2

*3 3 3 3


For each one, the number of permutations is easily computable using the number of arrangements.
Respectively :


*

*$\frac{4!}{3!1!} = 4$

*$\frac{4!}{2!1!1!} = 12$

*$\frac{4!}{2!2!} = 6$

*$\frac{4!}{1!2!1!} = 12$

*$\frac{4!}{4!} = 1$


So the sum is $35$.
