Discrete Probability: Uniformly random permutation of the set {1,2,3,..2n}












0














Question: Let $n geq 1$ be an integer. Consider a uniformly random permutation of the set ${1,2,3, ldots, 2n}$. Define the event:



A = "both the first element and the last element in the permutation are even integers"



What is $Pr(A)$?



Answer: $dfrac{(n-1)}{2(2n-1)}$



Attempt: I'm not sure how to go about this. I know there are $2$ positions out of $n$ positions so $nC2$ and there is $1$ way to have first and last element be even integers. And the $2n-2$ remaining integers we can arrange in $(2n-2)!$ ways. And this is out of $n!$ ways.



So, $Pr = dfrac{(nC2) cdot (2N-2)!}{n!}$



This approach doesn't give me the right answer.






probability combinatorics discrete-mathematics permutations







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    Please read this tutorial on how to typeset mathematics on this site.
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  • will do, sorry!
    – Toby
    Nov 27 '18 at 1:23
















0














Question: Let $n geq 1$ be an integer. Consider a uniformly random permutation of the set ${1,2,3, ldots, 2n}$. Define the event:



A = "both the first element and the last element in the permutation are even integers"



What is $Pr(A)$?



Answer: $dfrac{(n-1)}{2(2n-1)}$



Attempt: I'm not sure how to go about this. I know there are $2$ positions out of $n$ positions so $nC2$ and there is $1$ way to have first and last element be even integers. And the $2n-2$ remaining integers we can arrange in $(2n-2)!$ ways. And this is out of $n!$ ways.



So, $Pr = dfrac{(nC2) cdot (2N-2)!}{n!}$



This approach doesn't give me the right answer.






probability combinatorics discrete-mathematics permutations







share|cite|improve this question




















  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 27 '18 at 1:22










  • will do, sorry!
    – Toby
    Nov 27 '18 at 1:23














0












0








0







Question: Let $n geq 1$ be an integer. Consider a uniformly random permutation of the set ${1,2,3, ldots, 2n}$. Define the event:



A = "both the first element and the last element in the permutation are even integers"



What is $Pr(A)$?



Answer: $dfrac{(n-1)}{2(2n-1)}$



Attempt: I'm not sure how to go about this. I know there are $2$ positions out of $n$ positions so $nC2$ and there is $1$ way to have first and last element be even integers. And the $2n-2$ remaining integers we can arrange in $(2n-2)!$ ways. And this is out of $n!$ ways.



So, $Pr = dfrac{(nC2) cdot (2N-2)!}{n!}$



This approach doesn't give me the right answer.






probability combinatorics discrete-mathematics permutations







share|cite|improve this question















Question: Let $n geq 1$ be an integer. Consider a uniformly random permutation of the set ${1,2,3, ldots, 2n}$. Define the event:



A = "both the first element and the last element in the permutation are even integers"



What is $Pr(A)$?



Answer: $dfrac{(n-1)}{2(2n-1)}$



Attempt: I'm not sure how to go about this. I know there are $2$ positions out of $n$ positions so $nC2$ and there is $1$ way to have first and last element be even integers. And the $2n-2$ remaining integers we can arrange in $(2n-2)!$ ways. And this is out of $n!$ ways.



So, $Pr = dfrac{(nC2) cdot (2N-2)!}{n!}$



This approach doesn't give me the right answer.






probability combinatorics discrete-mathematics permutations




probability combinatorics discrete-mathematics permutations






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share|cite|improve this question













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edited Nov 27 '18 at 1:32









N. F. Taussig

43.5k93355




43.5k93355










asked Nov 27 '18 at 1:13









Toby

1577




1577








  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 27 '18 at 1:22










  • will do, sorry!
    – Toby
    Nov 27 '18 at 1:23














  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 27 '18 at 1:22










  • will do, sorry!
    – Toby
    Nov 27 '18 at 1:23








1




1




Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 '18 at 1:22




Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 27 '18 at 1:22












will do, sorry!
– Toby
Nov 27 '18 at 1:23




will do, sorry!
– Toby
Nov 27 '18 at 1:23










2 Answers
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There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is
$$Pr(A) = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$



Why is your answer incorrect?



You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.



You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set ${1, 2, 3, ldots, 2n}$. However, there are $(2n)!$ such arrangements.



With these corrections, you would have obtained
$$Pr(A) = frac{binom{n}{2} cdot 2!(2n - 2)!}{(2n)!} = frac{frac{n!}{2!(n - 2)!} cdot 2!(2n - 2)!}{(2n)!} = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$






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    1














    After we pick the two special even numbers, there are two options to arrange them.



    Also, remember that in the denominator, we have $2n$ items to permute.



    $$frac{binom{n}2cdot 2cdot (2n-2)!}{(2n)!}=frac{n(n-1)}{(2n)(2n-1)}=frac{n-1}{2(2n-1)}$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is
      $$Pr(A) = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$



      Why is your answer incorrect?



      You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.



      You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set ${1, 2, 3, ldots, 2n}$. However, there are $(2n)!$ such arrangements.



      With these corrections, you would have obtained
      $$Pr(A) = frac{binom{n}{2} cdot 2!(2n - 2)!}{(2n)!} = frac{frac{n!}{2!(n - 2)!} cdot 2!(2n - 2)!}{(2n)!} = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$






      share|cite|improve this answer




























        1














        There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is
        $$Pr(A) = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$



        Why is your answer incorrect?



        You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.



        You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set ${1, 2, 3, ldots, 2n}$. However, there are $(2n)!$ such arrangements.



        With these corrections, you would have obtained
        $$Pr(A) = frac{binom{n}{2} cdot 2!(2n - 2)!}{(2n)!} = frac{frac{n!}{2!(n - 2)!} cdot 2!(2n - 2)!}{(2n)!} = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$






        share|cite|improve this answer


























          1












          1








          1






          There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is
          $$Pr(A) = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$



          Why is your answer incorrect?



          You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.



          You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set ${1, 2, 3, ldots, 2n}$. However, there are $(2n)!$ such arrangements.



          With these corrections, you would have obtained
          $$Pr(A) = frac{binom{n}{2} cdot 2!(2n - 2)!}{(2n)!} = frac{frac{n!}{2!(n - 2)!} cdot 2!(2n - 2)!}{(2n)!} = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$






          share|cite|improve this answer














          There are $2n$ ways to choose the first number and $2n - 1$ ways to choose the last number. Of these, there are $n$ favorable choices for the first number since it must be even, and $n - 1$ favorable choices for the last number since it must be one of the remaining even numbers. Thus, the probability that both the first and last numbers are even is
          $$Pr(A) = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$



          Why is your answer incorrect?



          You selected two even numbers to place in the first and last positions, then asserted that there is one way to have the first and last elements be even integers. However, there are two ways to arrange the selected integers in those positions.



          You also asserted that there are $n!$ ways to arrange the $2n$ integers in the set ${1, 2, 3, ldots, 2n}$. However, there are $(2n)!$ such arrangements.



          With these corrections, you would have obtained
          $$Pr(A) = frac{binom{n}{2} cdot 2!(2n - 2)!}{(2n)!} = frac{frac{n!}{2!(n - 2)!} cdot 2!(2n - 2)!}{(2n)!} = frac{n(n - 1)}{2n(2n - 1)} = frac{n - 1}{2(2n - 1)}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 '18 at 1:28

























          answered Nov 27 '18 at 1:18









          N. F. Taussig

          43.5k93355




          43.5k93355























              1














              After we pick the two special even numbers, there are two options to arrange them.



              Also, remember that in the denominator, we have $2n$ items to permute.



              $$frac{binom{n}2cdot 2cdot (2n-2)!}{(2n)!}=frac{n(n-1)}{(2n)(2n-1)}=frac{n-1}{2(2n-1)}$$






              share|cite|improve this answer


























                1














                After we pick the two special even numbers, there are two options to arrange them.



                Also, remember that in the denominator, we have $2n$ items to permute.



                $$frac{binom{n}2cdot 2cdot (2n-2)!}{(2n)!}=frac{n(n-1)}{(2n)(2n-1)}=frac{n-1}{2(2n-1)}$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  After we pick the two special even numbers, there are two options to arrange them.



                  Also, remember that in the denominator, we have $2n$ items to permute.



                  $$frac{binom{n}2cdot 2cdot (2n-2)!}{(2n)!}=frac{n(n-1)}{(2n)(2n-1)}=frac{n-1}{2(2n-1)}$$






                  share|cite|improve this answer












                  After we pick the two special even numbers, there are two options to arrange them.



                  Also, remember that in the denominator, we have $2n$ items to permute.



                  $$frac{binom{n}2cdot 2cdot (2n-2)!}{(2n)!}=frac{n(n-1)}{(2n)(2n-1)}=frac{n-1}{2(2n-1)}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 1:17









                  Siong Thye Goh

                  99.4k1464117




                  99.4k1464117






























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