Distributing 3 Pears & 4 Apples to 9 People: Combinations

[tgt]

Homework Statement

If we like to distribute 3 pears and 4 apples to 9 people such that no 2 or more fruit is given to the same person, in how many ways can this be done?

The Attempt at a Solution

(9,7)[(7,3)-(6!/4!+5!/3!)] where (a,b) represents the 'a choose b' function.

 

[tiny-tim]

Hi tgt!

If we like to distribute 3 pears and 4 apples to 9 people such that no 2 or more fruit is given to the same person, in how many ways can this be done?

urgh! :yuck:

Hint: let's rewrite the question …

we want to distribute 3 pears and 4 apples and 2 nothings to 9 people so that they get one each.

 

[tgt]

Hi tgt!


urgh! :yuck:

Hint: let's rewrite the question …

we want to distribute 3 pears and 4 apples and 2 nothings to 9 people so that they get one each.

No. A person is allowed to have two pieces of fruit such as one pear and one apple.

 

[tiny-tim]

No. A person is allowed to have two pieces of fruit such as one pear and one apple.

Are you sure?

You originally quoted:

… no 2 or more fruit is given to the same person*…

which includes "no 2 fruit is given to the same person" …

in other words, one or none each.

 

[HallsofIvy]

tiny-tim, I would interpret "no 2 or more fruit" as meaning each person gets either 0 or 1 fruit of any kind- so "one pear and one apple" wou.ld violate that. And I would interpret "A gets an apple" as different from "A gets a pear". That's a noticeably harder problem.

 

[tgt]

It seems both HallsoIvy and tiny tim are correct. In which case the answer is 9!/(2!3!4!)

But that's not a noticeable hard problem, hallsofivy?