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Thread: Counting using inclusion/exclusion principle.

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    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

    $\displaystyle A_1$ = {selections of candy with exactly two pieces of type 1)
    $\displaystyle A_2$ = {selections of candy with exactly two pieces of type 2)
    $\displaystyle A_3$ = {selections of candy with exactly two pieces of type 3)
    $\displaystyle A_4$ = {selections of candy with exactly two pieces of type 4)


    the cardinality of the universe is $\displaystyle C(14,11)$ (11 dots, 3 slashes)

    For cardinality of $\displaystyle A_1$, 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 $\displaystyle n(A_1) =$ $\displaystyle C(11,9)$

    (note: this is also the cardinality of $\displaystyle A_2,A_3,A_4$)

    For intersections of two, there are 6 such intersections. They have cardinality $\displaystyle C(8,7)$. (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 $\displaystyle C(5,0)$ or 1.

    The intersection of all four is empty. Since 2+2+2+2 $\displaystyle \neq$ 11

    So via IEP

    $\displaystyle n(A_1 \cup A_2 \cup A_3 \cup A_4) = 4*C(11,9)-6*C(8,7)+4$

    subtract above from $\displaystyle n(U)$

    $\displaystyle C(14,11)-[4*C(11,9)-6*C(8,7)+4]$

    I was wondering if I did this correctly since the answer in the book does not have the $\displaystyle +4$ at the end.

    Thanks for helping me. Please let me know if I should clarify anything.
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    Re: Counting using inclusion/exclusion principle.

    Quote Originally Posted by Jame View Post
    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?
    $\displaystyle A_1$ = {selections of candy with exactly two pieces of type 1)
    $\displaystyle A_2$ = {selections of candy with exactly two pieces of type 2)
    $\displaystyle A_3$ = {selections of candy with exactly two pieces of type 3)
    $\displaystyle A_4$ = {selections of candy with exactly two pieces of type 4)
    For cardinality of $\displaystyle A_1$, 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 $\displaystyle n(A_1) =$ $\displaystyle C(11,9)$

    (note: this is also the cardinality of $\displaystyle A_2,A_3,A_4$)

    For intersections of two, there are 6 such intersections. They have cardinality $\displaystyle C(8,7)$. (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 $\displaystyle C(5,0)$ or 1.

    The intersection of all four is empty. Since 2+2+2+2 $\displaystyle \neq$ 11

    So via IEP

    $\displaystyle n(A_1 \cup A_2 \cup A_3 \cup A_4) = 4*C(11,9)-6*C(8,7)+4$

    subtract above from $\displaystyle n(U)$

    $\displaystyle C(14,11)-[4*C(11,9)-6*C(8,7)+4]$

    I was wondering if I did this correctly since the answer in the book does not have the $\displaystyle +4$ at the end.

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