Fruit Distribution Combinations

Question

(In this question all apples are considered indistinguishable, same for oranges and for peaches.)

1) In how many ways 7 apples, 8 oranges, and 9 peaches can be shared between 2 people?

I calculated this as C 24,2 = 276

2) In how many ways 7 apples, 8 oranges, and 9 peaches can be shared between 3 people?

I calcuated this as C 24,3 = 2024

Am I on the right track?

Answer 1

1)

Let $x_1$ and $x_2$ represent the number of apples given to the people. Hence we need non negative integral solutions of $x_1+x_2=7$ which is $8$ using star and bars method.

Let $x_3$ and $x_4$ represent the number of oranges given to the people. Hence we need non negative integral solutions of $x_3+x_4=8$ which is $9$.

Let $x_5$ and $x_6$ represent the number of peaches given to the people. Hence we need non negative integral solutions of $x_5+x_6=9$ which is $10$.

Hence final answer is $8*9*10=720$ 2)

Using the same intuition we find non negative integral solutions of

$y_1+y_2+y_3=7$ as $\binom {9}{2}$

$y_4+y_5+y_6=8$ as $\binom {10}{2}$

$y_7+y_8+y_9=9$ as $\binom {11}{2}$

Hence the final answer as $36*45*55=89100$

Answer 2

Choosing 2 from 24, $\binom {24}{2}$, doesn't seem to fit the requirements here. What are you choosing? Why two?

If there is no constraint on how much fruit each person has, then (in question 1) you can simply make a free choice for one person and the other gets whatever is left. So there are $8$ possible options for the number of apples, $0$ to $7$, and likewise $9$ choices for the oranges and $10$ choices for the peaches, giving $8\times 9\times 10 = 720$ total options.

If there is a constraint on the fruit so that each person has 12 items, I would probably run inclusion-exclusion on a couple of star-and-bars choices. so the first person, taking $12$ items from an unlimited supply of the $3$ fruits, would have $\binom {12+2}2 = 91$ options. However there are some of those that break the limits of fruit available, so we can count the excess apple cases by "preloading" 7 apples and counting the divisions of the remaining 5 fruits, $\binom {5+2}2=21$ and likewise oranges $\binom {4+2}2=15$ and peaches $\binom {3+2}2=10$ for a total of $91-21-15-10=45$ valid options in this case.

The second question is not too hard under the assumption that the fruit count doesn't have to be equal - just run stars-and-bars on each fruit separately and multiply - but gets complicated for keeping the fruit count equal.