How many ways are there to distribute seven distinct apples How many ways are there to distribute seven distinct apples and six distinct pears to 3 distinct people, such that each person has at least one pear? 
Attempt: So I broke it into two parts: Apples and the pears,  then I will sum them up.
Apples: $3^7$
Pears: $\binom{6}{3}.......$
So I chose C(6,3) because I thought "choosing 3 of the distinct pears to fill one per person,  then worry about how to allocate the other 3 pears"  So my issue is trying to figure out how to allocate the other pears...... 
 A: Note that the apples are distinct and the pears are distinct (e.g. if you have red delicious, golden delicious, granny smith, macintosh, pink lady, etc...)
As you say, you can break this into multiplication principle by first distributing the apples, and then distributing the pears and multiplying the results.
You correctly found $3^7$ as the number of ways of distributing the apples.  Now we ask how to distribute the pears such that everyone gets at least one.
To do this, approach via inclusion exclusion.  Let $A$ be the event that our first person receives no pears.  Let $B$ be the event that our second person receives no pears.  Let $C$ be the event that our third person receives no pears.  Let $U$ be universal event of all ways of distributing pears if we don't care about restrictions.
We are asked to count $|A^c\cap B^c\cap C^c| = |U\setminus(A\cup B\cup C)|$
This is, by inclusion exclusion:
$=|U| - |A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
Now., to calculate each of these: $|U| = 3^6$ as in the apples case
$|A|$ is the number of ways of distributing such that the first person gets none.  So, for each pear, they have two possible destinations.  $|A|=2^6$.  Similarly for $|B|$ and $|C|$.
$|A\cap B|$ is the number of ways such that person 1 and person two receive none.  There is only one way for this to occur.  Similarly for $|A\cap C|$ and $|B\cap C|$.
Finally, it is impossible for all three of the people to receive none of the pears.
There are then $3^6-3\cdot 2^6+3 = 540$ possible ways to distribute the pears.
Multiplying by the number of ways we could distribute the apples, we arrive at a total of $3^7\cdot 540 = 1180980$
A: Use inclusion/exclusion principle:

Include the number of ways to distribute them such that at most $\color\red3$ people have pears:
$$3^7\cdot\binom{3}{\color\red3}\cdot\color\red3^6=1594323$$

Exclude the number of ways to distribute them such that at most $\color\red2$ people have pears:
$$3^7\cdot\binom{3}{\color\red2}\cdot\color\red2^6=419904$$

Include the number of ways to distribute them such that at most $\color\red1$ person has pears:
$$3^7\cdot\binom{3}{\color\red1}\cdot\color\red1^6=6561$$

Hence the number of ways to distribute them such that each person has pears is:
$$1594323-419904+6561=1180980$$
A: Hint: Use the Stirling number of the second kind to calculate the numbers of ways to distribute the pears.
$$ n!\cdot S_{k,n}=n!\cdot\frac{1}{n!}\cdot \sum_{j=0}^n (-1)^{n-j} {n \choose j} j^k $$
k is the number of pears (balls). n is the number of people (urns).
A: I am just doing it for pears.
This becomes very simple when you notice this is nothing but indirect application of ONTO function. Every element in range must have a pre-image(Therefore range and co-domain both are equivalent for onto functions).
Here we are distributing distinct pears(which becomes domain in this case) to distinct people(which becomes range in this case) such that each people must have at least one pear.
You must have idea how to calculate number of onto functions else google and read about it. Thats it! 
The same answer $540$ will come in this process too. This is just a part of solution because you need to multiply it to $3^7$ to get the final answer.
