Number of ways to distribute 4 different objects and 5 identical objects in 3 separate groups? So, the question goes as:
The number of ways in which 4 different toys and 5 identical marbles can be distributed between 3 different people, if each person gets at least one toy and one marble is?
My attempt:
I started off by distributing 3 marbles to each person. There's only 1 way to do it, since marbles are identical. Then,I distributed the rest of the two marbles. First off, each guy gets both the marbles. That makes for 3 ways. Then 2 guys get a marble each. This too makes for 3 ways.
So, total number of ways to distribute marbles is 6.
Now, to distribution of toys. First, I pick 3 toys out of 4 and distribute them. This takes $^4P_3$ methods to do. Now remains the last toy. There are 3 people, and hence 3 ways to distribute it.
Now, the total goes to:
$6× ^4P_3×3$ ways.
This comes out to be 532.
The Problem:
I can't seem to find fault in my logic, but the answer key says the answer is 216. 
Can anyone please point out what I am doing wrong?
 A: You have overcounted the number of ways to distribute the toys. Suppose the toys are $T_1,T_2,T_3,T_4$. Then one of the toy distributions you counted is where you pick out $T_1,T_2,T_3$ and distribute them to person 1, person 2, and person 3, in that order, and give $T_4$ to person 1. But you've also counted, as a separate way of distributing the toys, the distribution where you pick out $T_2,T_3,T_4$ and give $T_2$ to person 2, $T_4$ to person 3, and $T_4$ to person 1, and then give $T_1$ to person 1. This is, however, the same toy distribution as the previous one. 
You can count the number of ways to distribute the toys as follows. Exactly one person must receive two toys, and there are $3$ possibilities for who that person is. There are ${4 \choose 2}=6$ choices for which toys that person receives. There are two remaining toys and two people who must each receive one of them, so there are $2$ ways to distribute the last two toys. This gives $3 \cdot 6 \cdot 2 = 36$ ways to distribute the toys.
Now you can multiply as you did before to get $6 \cdot 36=216$ ways to distribute the marbles and toys.
A: Note that $6 \cdot P(4, 3) \cdot 3 = \color{red}{432}$.  Your answer is twice the actual answer.  
Why?
Your method counts each assignment of toys twice, once for each order in which the person who receives two toys can receive those two toys.  
To illustrate, suppose you give the first person the first toy, the second person the second toy, and the third person the third and fourth toys.  Your method counts this assignment of toys twice, once when the third person receives the third toy then the fourth toy and once when the third person receives the fourth toy then the third toy.  Hence, the number of ways of distributing the toys is 
$$\frac{3 \cdot P(4, 3)}{2} = \frac{3 \cdot 24}{2} = 3 \cdot 12 = 36$$
Since you are correct that there are six ways of distributing the marbles, there are 
$$6 \cdot 36 = 216$$
ways of distributing the marbles and toys to the three people in such a way that each person receives at least one marble and at least one toy.
Alternate Method:  Let $x_k$ represent the number of marbles given to the $k$th person, then 
$$x_1 + x_2 + x_3 = 5$$
which, since each person receives at least one marble, is an equation in the positive integers.  A particular solution corresponds to placing two addition signs in the four spaces between the ones in a row of five ones.  For instance, 
$$1 1 + 1 1 + 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 2$, and $x_3 = 1$, while 
$$1 1 1 + 1 + 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 1$, and $x_3 = 1$.  Thus, the number of solutions is the number of ways two addition signs can be placed in the four spaces between five successive ones, which is 
$$\binom{4}{2} = 6$$
If there were no restrictions on distributing the toys, we would have three ways to distribute each of the four toys, so there would be $3^4$ ways to distribute the toys.  We must exclude from these those distributions in which not all the people receive a toy.  
There are $\binom{3}{1}$ ways of selecting a person not to receive a toy and $2^4$ ways of distributing the four toys to two people. There are $\binom{3}{2}$ ways of selecting two people not to receive a toy and $1^4$ ways of giving all four toys to one person.  
By the Inclusion-Exclusion Principle, the number of ways the four toys can be distributed so that all three people receive at least one toy is 
$$3^4 - \binom{3}{1}2^4 + \binom{3}{2}1^4 = 81 - 3 \cdot 16 + 3 \cdot 1 = 81 - 48 + 3 = 36$$
Hence, the number of ways of distributing five identical marbles and four different toys to three people so that each person receives at least one marble and at least one toy is 
$$\binom{4}{2}\left[3^4 - \binom{3}{1}2^4 + \binom{3}{2}1^4\right] = 6 \cdot 36 = 216$$
as we found above.
