In how many ways can $6$ chocolates and $6$ popcorns be given to $10$ girls so that every girl receives at least one item? In how many ways can $6$ chocolates and $6$ popcorns be given to $10$ girls such that each girl receives at least one of either chocolates or popcorns?
I know that if it would have been only $6$ chocolates given to $10$ girls, then it is straightforward problem (stars and bars).
Is there any shorter way to approach the above problem?
 A: We will distribute the chocolates first, then the popcorn.  To ensure that each girl receives at least one item, we must distribute the chocolates in such a way that at least four girls receive a piece of chocolate, otherwise at least one girl would receive nothing.
Exactly four girls receive a piece of chocolate:  There are $\binom{10}{4}$ ways to choose which four girls receive a piece of chocolate.  Hand each of them one piece of chocolate.  Hand each of the other six girls one container of popcorn.  That leaves us with two pieces of chocolate to distribute to the four girls who have received a piece of chocolate.  The number of ways this can be done is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 2$$
in the nonnegative integers, which is 
$$\binom{2 + 4 - 1}{4 - 1} = \binom{5}{3}$$
Hence, there are 
$$\binom{10}{4}\binom{5}{3}$$
ways to distribute six pieces of chocolates and six containers of popcorn to ten girls if each girl receives at least one chocolate or at least one popcorn and exactly four girls receive a piece of chocolate.
Exactly five girls receive a piece of chocolate:  There are $\binom{10}{5}$ ways to select which five of the ten girls will receive a piece of chocolate.  Give each of these girls a piece of chocolate.  Give each girl who did not receive a chocolate a container of popcorn.  This leaves us with five ways to distribute the remaining piece of chocolate to one of the five girls selected to receive a piece of chocolate.  The remaining container of popcorn can be distributed to any of the ten girls.  Hence, there are 
$$\binom{10}{5}\binom{5}{1}\binom{10}{1}$$
ways to distribute six pieces of chocolate and six pieces of popcorn to ten girls if each girl receives a piece of chocolate or a container of popcorn and exactly five girls receive a piece of chocolate.
Exactly six girls receive a piece of chocolate:  There are $\binom{10}{6}$ ways to choose which six girls receive a piece of chocolate.  Hand each of the other four girls a container of popcorn.  That leaves two containers of popcorn to distribute to ten girls.  The number of ways this can be done is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} = 2$$
in the nonnegative integers, which is 
$$\binom{2 + 10 - 1}{10 - 1} = \binom{11}{9}$$
Hence, the number of ways six pieces of chocolate and six containers of popcorn can be distributed to ten girls if each girl receives a piece of chocolate or a container of popcorn and exactly six girls receive a piece of chocolate is 
$$\binom{10}{6}\binom{11}{9}$$
Total:  Since the cases above are mutually exclusive and exhaustive, the total can be found by adding the above cases.  
A: Let's distribute the $6$ chocolates first. Since at least $4$ girls have to receive one of them we have to list the partitions of $6$ into $\geq4$ parts. They are
$$(3,1,1,1),\quad (2,2,1,1),\quad (2,1,1,1,1),\quad (1,1,1,1,1,1)\ .$$
 In case $(3,1,1,1)$ we can choose the girl receiving $3$ chocolates in ${10\choose1}$ ways and then the three girls receiving $1$ chocolate in ${9\choose 3}$ ways. In case $(2,2,1,1)$ we can choose the two girls receiving $2$ chocolates in ${10\choose2}$ ways and then the two girls receiving $1$ chocolate in ${8\choose 2}$ ways. In both these  cases each of the remaining $6$ girls obtains a single popcorn. Makes
$${10\choose1}{9\choose3}+{10\choose2}{8\choose 2}$$
allocations so far.
In case $(2,1,1,1,1)$ we can choose the girl receiving $2$ chocolates in ${10\choose1}$ ways and then the $4$ girls receiving $1$ chocolate in ${9\choose4}$ ways. Now we need $5$ popcorns to satisfy the remaining $5$ girls, and there is one popcorn left over to give to any of the $10$ girls. Makes
$${10\choose1}{9\choose4}{10\choose1}$$ possible allocations.
In case $(1,1,1,1,1,1)$ we can choose the $6$ girls receiving a chocolate in ${10\choose6}$ ways and then spend $4$ popcorns on the remaining $4$ girls. There remain two popcorns which we can give to two girls in ${10\choose2}$ ways or to a single girl in ${10\choose1}$ ways. Makes
$${10\choose6}\left({10\choose2}+{10\choose1}\right)$$
admissible allocations.
Adding it all up brings us to $26\,250$ possibilities.
A: I shall focus on making blocks when a child receives more than one piece
Suppose we give $3$ chocolates to one kid: $\boxed{CCC},  3C's, 6P's$ are left to distribute to remaining $9$,  and ways this can happen is $\binom{10}1\binom93$, with a mirror image for giving $3$ popcorns to one kid$\;\boxed{PPP}$
With the basic method as above, we get:


*

*$\boxed{CCC}\;or\; \boxed{PPP}: 2\binom{10}1\binom93 = 1680$

*$\boxed{CCP}\;or\; \boxed{PPC}: 2\binom{10}1\binom94 = 2520$

*$\boxed{CC}\;and\; \boxed{PP}: \binom{10}2 2!\binom84 = 6300$

*$\boxed{CP}\;and\; \boxed{CP}: \binom{10}2\binom84 = 3150$

*$\boxed{CC}\;and\; \boxed{CP}\; or\; \boxed{PP}\;and\;\boxed{CP}: 2\binom{10}2 2!\binom83 = 10080$

*$\boxed{CC}\;and\;\boxed{CC}\;or\; \boxed{PP}\;and\;\boxed{PP}: 2\binom{10}2\binom82 = 2520$
The total comes to $\boxed{26250}$
