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.