bags of chocolate problem I have $4$ different types of chocolates. How many unique bags of chocolate can I make with $10$ items per bag that has at least one type of each chocolate in each bag?
I don't know if this is correct but this is what i got:
$$\frac{\frac{10!}{6!}\cdot\frac{6!}{4!}\cdot\frac{4!}{2!}}{10!}$$
 A: Assuming each piece of chocolate has an infinite repetition number or at least $7$ pieces of each type we can consider $$c_1+c_2+c_3+c_4=10$$ where $1<c_i<7$ for each $i\in\{1,2,3,4\}$. We make the substitution $x_i=c_1-1$ and we obtain the equation $$x_1+x_2+x_3+x_4=6$$ where $0<x_i<6$ for each $i\in\{1,2,3,4\}$. The number of non-negative integral solutions to this equation is ${6+4-1\choose 6}={9\choose 6}=84$. Thus there are $84$ unique bags of chocolate where each bag contains at least one piece of each type of chocolate.
A: Hint:
Give the types the names c1, c2, c3 and c4 and start by putting
of each of the types a piece of chocolate in the bag. Then $6$ pieces
of chocolate are to be added. That means that somehow number $6$
must be split up in $4$ parts. For instance: $6=2+0+3+1$ stands for
the possibility that $2$ pieces are taken from c1, $0$ from c2,
$3$ from type c3 and $1$ from c4. You must find out in how many
ways $6$ can be split up.
Additional hint:
Start with $6+4-1=9$ zeros on a row: $000000000$. Pick $3$ zeros
out and change them into a $1$. For instance: $001100010$. This
gives the split up $6=2+0+3+1$ in the sense: $2$ zeros on the left
of the first $1$; $0$ zeros between the first and the second $1$;
$3$ zeros between the second and the third $1$ and finally $1$
zero at the right of third $1$. This shows that splitting up $6$
in $4$ parts can be interpreted as choosing $3$ out of $9$.
