Combinatorics problem Algebra












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Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$
but how can i calculate them without writing them down?










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  • You mean noone has his own ball?
    – greedoid
    Nov 18 at 10:11










  • Yes, not the one that he had at the start
    – Sdwae
    Nov 18 at 10:13
















0














Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$
but how can i calculate them without writing them down?










share|cite|improve this question






















  • You mean noone has his own ball?
    – greedoid
    Nov 18 at 10:11










  • Yes, not the one that he had at the start
    – Sdwae
    Nov 18 at 10:13














0












0








0


1





Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$
but how can i calculate them without writing them down?










share|cite|improve this question













Four friends have four different balls and they decide to swap them so that everyone has a different ball. How many possible combinations are there?
I know they are $9$: $2, 1, 4, 3 - 2, 3, 4, 1 - 2, 4, 1, 3 - 3, 1, 4, 2 -
3, 4, 1, 2 - 3, 4, 2, 1 - 4, 1, 2, 3 - 4, 3, 1, 2 - 4, 3, 2, 1$
but how can i calculate them without writing them down?







combinatorics






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asked Nov 18 at 10:08









Sdwae

52




52












  • You mean noone has his own ball?
    – greedoid
    Nov 18 at 10:11










  • Yes, not the one that he had at the start
    – Sdwae
    Nov 18 at 10:13


















  • You mean noone has his own ball?
    – greedoid
    Nov 18 at 10:11










  • Yes, not the one that he had at the start
    – Sdwae
    Nov 18 at 10:13
















You mean noone has his own ball?
– greedoid
Nov 18 at 10:11




You mean noone has his own ball?
– greedoid
Nov 18 at 10:11












Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13




Yes, not the one that he had at the start
– Sdwae
Nov 18 at 10:13










1 Answer
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These are called derangements and have been well studied. For any $n$ there are $$
n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
$$






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    1 Answer
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    1 Answer
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    active

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    active

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    active

    oldest

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    1














    These are called derangements and have been well studied. For any $n$ there are $$
    n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
    $$






    share|cite|improve this answer


























      1














      These are called derangements and have been well studied. For any $n$ there are $$
      n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
      $$






      share|cite|improve this answer
























        1












        1








        1






        These are called derangements and have been well studied. For any $n$ there are $$
        n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
        $$






        share|cite|improve this answer












        These are called derangements and have been well studied. For any $n$ there are $$
        n!Sigma_{i=0}^nfrac{(-1)^i}{i!}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 18 at 10:11









        Joey Kilpatrick

        1,181422




        1,181422






























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