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Math Help - [SOLVED] SETS AND FUNCTIONS

  1. #1
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    [SOLVED] SETS AND FUNCTIONS


    Let S be the set of all functions from {1, 2} to {1, 2, 3}. Let F be an element of S chosen at random.

    Find |S| and find the probability that F is an injective function.


    For the question I know the |S|= 9 but finding it difficult to find the probability that f is an injective function.

    Let S be the set of all functions from {1, 2, 3} to {1, 2}. Let G be an element of S chosen at random.

    Find |S| and find the probability that G is an injective function.

    For the question I know that the probability that f is an injective function is 0 as this can not be injective as the domain and co domain is disjoint. But I am finding it difficult to find |S|.

    Let S be the set of all functions from {1, 2,Ö, M} to {1, 2,Ö, N}. Let Q be an element of S chosen at random.


    Find |S| and find the probability that Q is an injective function.

    For this question I do not understand it at all.
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  2. #2
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    In order for there to be an injection from A \mapsto B it must be the case that <br />
\left| A \right| \leqslant \left| B \right|.
    Then we count the number of permutations of  \left| B \right| taken  \left| A \right| at a time: \frac{{\left| B \right|!}}{{\left[ {\left| B \right| - \left| A \right|} \right]!}}.
    In terms of probability <br />
\frac{{\frac{{\left| B \right|!}}{{\left[ {\left| B \right| - \left| A \right|} \right]!}}}}{{\left| B \right|^{\left| A \right|} }}.
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  3. #3
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    Quote Originally Posted by Plato View Post
    In order for there to be an injection from A \mapsto B it must be the case that <br />
\left| A \right| \leqslant \left| B \right|.
    Then we count the number of permutations of  \left| B \right| taken  \left| A \right| at a time: \frac{{\left| B \right|!}}{{\left[ {\left| B \right| - \left| A \right|} \right]!}}.
    In terms of probability <br />
\frac{{\frac{{\left| B \right|!}}{{\left[ {\left| B \right| - \left| A \right|} \right]!}}}}{{\left| B \right|^{\left| A \right|} }}.
    Thanks for for the help. I do understand the first part and what is meant by injective but i do not understand about the following:

    Then we count the number of permutations of taken at a time: .
    In terms of probability and the question.

    Can you please explain it to me in more detail.

    Thanks

    The1u2001
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  4. #4
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    Letís say that \left| A \right| = 5\;\& \;\left| B \right| = 9 so there are 9^5\ = 59049 possible mappings A\mapsto B.
    The set B^A, the set of all mappings A\mapsto B, has cardinality \left| {B^A } \right| = \left| B \right|^{\left| A \right|}. (The notation makes it easy to remember.)

    Now the number of injections A\mapsto B is P(9,5) = \frac{{9!}}{{\left[ {9 - 5} \right]!}} = \left( 9 \right)\left( 8 \right)\left( 7 \right)\left( 6 \right)\left( 5 \right) = 15120.

    So the probability of randomly selecting an injection is \frac {15120}{59049}.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Letís say that \left| A \right| = 5\;\& \;\left| B \right| = 9 so there are 9^5\ = 59049 possible mappings A\mapsto B.
    The set B^A, the set of all mappings A\mapsto B, has cardinality \left| {B^A } \right| = \left| B \right|^{\left| A \right|}. (The notation makes it easy to remember.)

    Now the number of injections A\mapsto B is P(9,5) = \frac{{9!}}{{\left[ {9 - 5} \right]!}} = \left( 9 \right)\left( 8 \right)\left( 7 \right)\left( 6 \right)\left( 5 \right) = 15120.

    So the probability of randomly selecting an injection is \frac {15120}{59049}.
    Know i get it.

    Once again, thanks for helping me, i really appriate it.

    THE1U2001
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