Thread: Counting using inclusion/exclusion principle.

Hello, I did this problem and wanted to know if I was correct since the answer in the back the book is different.

Question:

In how many ways can a child take 11 candies where there are four types of candies if the child does not take exactly 2 pieces of any type?

I will use inclusion/exclusion. Let

= {selections of candy with exactly two pieces of type 1)
= {selections of candy with exactly two pieces of type 2)
= {selections of candy with exactly two pieces of type 3)
= {selections of candy with exactly two pieces of type 4)


the cardinality of the universe is (11 dots, 3 slashes)

For cardinality of , we put two candies in type 1 and we have to distribute the remaining 9 candies into the other three types. (9 dots, 2 slashes)

so

(note: this is also the cardinality of )

For intersections of two, there are 6 such intersections. They have cardinality . (i.e. we have exactly two of two types, distribute the remaining 7 candies among the two other types. 7 dots, 1 slash)

for intersections of three, there are four such intersections.

Fix two candies in three types, place remaining 5 in fourth type.

the cardinality is or 1.

The intersection of all four is empty. Since 2+2+2+2 11

So via IEP



subtract above from



I was wondering if I did this correctly since the answer in the book does not have the at the end.

Thanks for helping me. Please let me know if I should clarify anything.

Originally Posted by Jame

Hello, I did this problem and wanted to know if I was correct since the answer in the back the book is different.
Question: In how many ways can a child take 11 candies where there are four types of candies if the child does not take exactly 2 pieces of any type?
= {selections of candy with exactly two pieces of type 1)
= {selections of candy with exactly two pieces of type 2)
= {selections of candy with exactly two pieces of type 3)
= {selections of candy with exactly two pieces of type 4)
For cardinality of , we put two candies in type 1 and we have to distribute the remaining 9 candies into the other three types. (9 dots, 2 slashes)

so

(note: this is also the cardinality of )

For intersections of two, there are 6 such intersections. They have cardinality . (i.e. we have exactly two of two types, distribute the remaining 7 candies among the two other types. 7 dots, 1 slash) for intersections of three, there are four such intersections.

Fix two candies in three types, place remaining 5 in fourth type.

the cardinality is or 1.

The intersection of all four is empty. Since 2+2+2+2 11

So via IEP



subtract above from



I was wondering if I did this correctly since the answer in the book does not have the at the end.

Thanks for helping me. Please let me know if I should clarify anything.

You are correct.