0
$\begingroup$

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:

$$1594323-419904+6561=1180980$$

$\endgroup$
  • $\begingroup$ This number is far too large. You are, I think, treating the various apples (say) as distinguishable. There aren't, for example, $3^7$ ways to distribute the apples, far from it. Hint: apply Stars and Bars to each type of fruit separately. $\endgroup$ – lulu Feb 20 '18 at 19:26
  • $\begingroup$ @lulu thanks, could you help me through this? $\endgroup$ – fr14 Feb 20 '18 at 19:29
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I see where I went wrong, thanks! $\endgroup$ – fr14 Feb 20 '18 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.