How many ways are there to distribute $7$ (identical) apples, $6$ oranges and $7$ pears among $3$ different people with each person getting at least $1$ pear?
Below are my workings but I am not sure if they are correct, looking for some help, thanks!
Using the 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:
$$1594323-419904+6561=1180980$$
Best Answer
It's important to recognize that the fruit (of a given type) are meant to be indistinguishable. Thus, factors like $3^7$ should not appear.
We will use Stars and Bars:
For the pears, we need the number of $3-$tuples of positive integers that add to $7$. That's $$\binom 62=15$$
For the apples, we need the number of $3-$tuples of non-negative integers that add to $7$. That's $$\binom {7+3-1}{7}=36$$
For the oranges, we need the number of $3-$tuples of non-negative integers that add to $6$. That's $$\binom {6+3-1}{6}=28$$
Finally, the answer is the product $$\boxed {15\times 36\times 28 = 15120}$$
Note: as always with combinatorics, it's always a good idea to check the arithmetic carefully.