$1,2, ldots, n$ are permuted. None of the numbers $1,2,3$ are adjacent and $n>4$.












1














The numbers $1, 2, ldots, n$ are permuted. How many different permutations exist such that none of the numbers $1, 2, 3$ are adjacent when $n>4$?



Solution:
$4,5, ldots, n$ can be shuffled in $(n-3)!$ ways and $3!$ ways to arrange $1,2,3$. There are $n−2$ slots that are separated by $n−3$ shuffled numbers, and if we insert each of $1,2,3$ into a different slot, they cannot be adjacent. There are $binom{n - 2}{3}$ ways to do this.



Why is it $binom{n-2}{3}$ ways? I don't get the explanation.










share|cite|improve this question
























  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Dec 1 at 10:02










  • Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
    – drhab
    Dec 1 at 10:09










  • Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
    – Shubham Johri
    Dec 1 at 10:23
















1














The numbers $1, 2, ldots, n$ are permuted. How many different permutations exist such that none of the numbers $1, 2, 3$ are adjacent when $n>4$?



Solution:
$4,5, ldots, n$ can be shuffled in $(n-3)!$ ways and $3!$ ways to arrange $1,2,3$. There are $n−2$ slots that are separated by $n−3$ shuffled numbers, and if we insert each of $1,2,3$ into a different slot, they cannot be adjacent. There are $binom{n - 2}{3}$ ways to do this.



Why is it $binom{n-2}{3}$ ways? I don't get the explanation.










share|cite|improve this question
























  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Dec 1 at 10:02










  • Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
    – drhab
    Dec 1 at 10:09










  • Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
    – Shubham Johri
    Dec 1 at 10:23














1












1








1







The numbers $1, 2, ldots, n$ are permuted. How many different permutations exist such that none of the numbers $1, 2, 3$ are adjacent when $n>4$?



Solution:
$4,5, ldots, n$ can be shuffled in $(n-3)!$ ways and $3!$ ways to arrange $1,2,3$. There are $n−2$ slots that are separated by $n−3$ shuffled numbers, and if we insert each of $1,2,3$ into a different slot, they cannot be adjacent. There are $binom{n - 2}{3}$ ways to do this.



Why is it $binom{n-2}{3}$ ways? I don't get the explanation.










share|cite|improve this question















The numbers $1, 2, ldots, n$ are permuted. How many different permutations exist such that none of the numbers $1, 2, 3$ are adjacent when $n>4$?



Solution:
$4,5, ldots, n$ can be shuffled in $(n-3)!$ ways and $3!$ ways to arrange $1,2,3$. There are $n−2$ slots that are separated by $n−3$ shuffled numbers, and if we insert each of $1,2,3$ into a different slot, they cannot be adjacent. There are $binom{n - 2}{3}$ ways to do this.



Why is it $binom{n-2}{3}$ ways? I don't get the explanation.







combinatorics permutations






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share|cite|improve this question













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share|cite|improve this question








edited Dec 1 at 10:01









N. F. Taussig

43.5k93355




43.5k93355










asked Dec 1 at 9:47









muffin

62




62












  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Dec 1 at 10:02










  • Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
    – drhab
    Dec 1 at 10:09










  • Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
    – Shubham Johri
    Dec 1 at 10:23


















  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Dec 1 at 10:02










  • Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
    – drhab
    Dec 1 at 10:09










  • Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
    – Shubham Johri
    Dec 1 at 10:23
















Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 1 at 10:02




Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 1 at 10:02












Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
– drhab
Dec 1 at 10:09




Mentioned is that from the $n-2$ slots $3$ are selected to be the slots where exactly one of the numbers $1,2,3$ is inserted. There are $binom{n-2}3$ to select $3$ out of $n-2$, right? Btw, after this observation we are not ready yet with the whole calculation.
– drhab
Dec 1 at 10:09












Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
– Shubham Johri
Dec 1 at 10:23




Have a look at en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
– Shubham Johri
Dec 1 at 10:23










4 Answers
4






active

oldest

votes


















0














Once you arrange $4,5,...,n$ in order, you have to put each of $1$, $2$ and $3$ either in a gap between two other numbers, or at one end of the list. But if you put one of them in a particular gap, or at a particular end, neither of the others can be put in the same place, or they would be adjacent in the final list.



Therefore you need to choose three different places to put $1,2,3$. There are $n-4$ gaps between the other numbers, and two additional places at the end, so $n-2$ places to choose from. The number of ways to choose $3$ things out of $n-2$ is $binom{n-2}3$.






share|cite|improve this answer





























    0














    You have $n-3$ numbers leaving $1,2,3$. Imagine you have a row of boxes, where each box denotes one of the $n-3$ numbers. You also have $n-2$ gaps: $n-4$ gaps between the boxes, and $1$ on each end.



    $$_ b_1 _ b_2 _ ... _ b_{n-3} _$$



    You have to permute the numbers in a manner such that $1,2,3$ are not adjacent to each other. This can be achieved if you select any $3$ distinct gaps out of the $n-2$ gaps to place one of $1,2,3$, since between any two gaps you have one or more boxes, or one or more numbers from the set ${4,5...,n}$, which ensures that $1,2,3$ are never adjacent to each other.



    You can select $3$ gaps in $binom{n-2}3$ ways. The numbers in those gaps can permute in $3!$ ways and the remaining numbers can permute in $(n-3)!$ ways, giving you a total of $(n-3)!cdot3!cdotbinom{n-2}3$ ways.






    share|cite|improve this answer





























      0














      There is a permutation of $1,2,3$ giving factor $3!$



      There is a permutation of $4,5,dots n$ giving factor $(n-3)!$



      The numbers $4,5,dots,n$ are "separated" (e.g. like $star7star5star4star6star$ if $n=7$) and $n-2$ slots/stars exists.



      These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.



      So we must select $3$ of $n-2$ stars giving factor $binom{n-2}3$.



      Then we end up with a total of: $$3!(n-3)!binom{n-2}3$$possibilities.






      share|cite|improve this answer





























        0














        We can count the cases




        • the permutations are n!


        from which we need to eliminate the ways




        • we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$

        • we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$






        share|cite|improve this answer























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          4 Answers
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          4 Answers
          4






          active

          oldest

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          active

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          active

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          0














          Once you arrange $4,5,...,n$ in order, you have to put each of $1$, $2$ and $3$ either in a gap between two other numbers, or at one end of the list. But if you put one of them in a particular gap, or at a particular end, neither of the others can be put in the same place, or they would be adjacent in the final list.



          Therefore you need to choose three different places to put $1,2,3$. There are $n-4$ gaps between the other numbers, and two additional places at the end, so $n-2$ places to choose from. The number of ways to choose $3$ things out of $n-2$ is $binom{n-2}3$.






          share|cite|improve this answer


























            0














            Once you arrange $4,5,...,n$ in order, you have to put each of $1$, $2$ and $3$ either in a gap between two other numbers, or at one end of the list. But if you put one of them in a particular gap, or at a particular end, neither of the others can be put in the same place, or they would be adjacent in the final list.



            Therefore you need to choose three different places to put $1,2,3$. There are $n-4$ gaps between the other numbers, and two additional places at the end, so $n-2$ places to choose from. The number of ways to choose $3$ things out of $n-2$ is $binom{n-2}3$.






            share|cite|improve this answer
























              0












              0








              0






              Once you arrange $4,5,...,n$ in order, you have to put each of $1$, $2$ and $3$ either in a gap between two other numbers, or at one end of the list. But if you put one of them in a particular gap, or at a particular end, neither of the others can be put in the same place, or they would be adjacent in the final list.



              Therefore you need to choose three different places to put $1,2,3$. There are $n-4$ gaps between the other numbers, and two additional places at the end, so $n-2$ places to choose from. The number of ways to choose $3$ things out of $n-2$ is $binom{n-2}3$.






              share|cite|improve this answer












              Once you arrange $4,5,...,n$ in order, you have to put each of $1$, $2$ and $3$ either in a gap between two other numbers, or at one end of the list. But if you put one of them in a particular gap, or at a particular end, neither of the others can be put in the same place, or they would be adjacent in the final list.



              Therefore you need to choose three different places to put $1,2,3$. There are $n-4$ gaps between the other numbers, and two additional places at the end, so $n-2$ places to choose from. The number of ways to choose $3$ things out of $n-2$ is $binom{n-2}3$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 1 at 10:12









              Especially Lime

              21.6k22858




              21.6k22858























                  0














                  You have $n-3$ numbers leaving $1,2,3$. Imagine you have a row of boxes, where each box denotes one of the $n-3$ numbers. You also have $n-2$ gaps: $n-4$ gaps between the boxes, and $1$ on each end.



                  $$_ b_1 _ b_2 _ ... _ b_{n-3} _$$



                  You have to permute the numbers in a manner such that $1,2,3$ are not adjacent to each other. This can be achieved if you select any $3$ distinct gaps out of the $n-2$ gaps to place one of $1,2,3$, since between any two gaps you have one or more boxes, or one or more numbers from the set ${4,5...,n}$, which ensures that $1,2,3$ are never adjacent to each other.



                  You can select $3$ gaps in $binom{n-2}3$ ways. The numbers in those gaps can permute in $3!$ ways and the remaining numbers can permute in $(n-3)!$ ways, giving you a total of $(n-3)!cdot3!cdotbinom{n-2}3$ ways.






                  share|cite|improve this answer


























                    0














                    You have $n-3$ numbers leaving $1,2,3$. Imagine you have a row of boxes, where each box denotes one of the $n-3$ numbers. You also have $n-2$ gaps: $n-4$ gaps between the boxes, and $1$ on each end.



                    $$_ b_1 _ b_2 _ ... _ b_{n-3} _$$



                    You have to permute the numbers in a manner such that $1,2,3$ are not adjacent to each other. This can be achieved if you select any $3$ distinct gaps out of the $n-2$ gaps to place one of $1,2,3$, since between any two gaps you have one or more boxes, or one or more numbers from the set ${4,5...,n}$, which ensures that $1,2,3$ are never adjacent to each other.



                    You can select $3$ gaps in $binom{n-2}3$ ways. The numbers in those gaps can permute in $3!$ ways and the remaining numbers can permute in $(n-3)!$ ways, giving you a total of $(n-3)!cdot3!cdotbinom{n-2}3$ ways.






                    share|cite|improve this answer
























                      0












                      0








                      0






                      You have $n-3$ numbers leaving $1,2,3$. Imagine you have a row of boxes, where each box denotes one of the $n-3$ numbers. You also have $n-2$ gaps: $n-4$ gaps between the boxes, and $1$ on each end.



                      $$_ b_1 _ b_2 _ ... _ b_{n-3} _$$



                      You have to permute the numbers in a manner such that $1,2,3$ are not adjacent to each other. This can be achieved if you select any $3$ distinct gaps out of the $n-2$ gaps to place one of $1,2,3$, since between any two gaps you have one or more boxes, or one or more numbers from the set ${4,5...,n}$, which ensures that $1,2,3$ are never adjacent to each other.



                      You can select $3$ gaps in $binom{n-2}3$ ways. The numbers in those gaps can permute in $3!$ ways and the remaining numbers can permute in $(n-3)!$ ways, giving you a total of $(n-3)!cdot3!cdotbinom{n-2}3$ ways.






                      share|cite|improve this answer












                      You have $n-3$ numbers leaving $1,2,3$. Imagine you have a row of boxes, where each box denotes one of the $n-3$ numbers. You also have $n-2$ gaps: $n-4$ gaps between the boxes, and $1$ on each end.



                      $$_ b_1 _ b_2 _ ... _ b_{n-3} _$$



                      You have to permute the numbers in a manner such that $1,2,3$ are not adjacent to each other. This can be achieved if you select any $3$ distinct gaps out of the $n-2$ gaps to place one of $1,2,3$, since between any two gaps you have one or more boxes, or one or more numbers from the set ${4,5...,n}$, which ensures that $1,2,3$ are never adjacent to each other.



                      You can select $3$ gaps in $binom{n-2}3$ ways. The numbers in those gaps can permute in $3!$ ways and the remaining numbers can permute in $(n-3)!$ ways, giving you a total of $(n-3)!cdot3!cdotbinom{n-2}3$ ways.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 at 10:16









                      Shubham Johri

                      3,861716




                      3,861716























                          0














                          There is a permutation of $1,2,3$ giving factor $3!$



                          There is a permutation of $4,5,dots n$ giving factor $(n-3)!$



                          The numbers $4,5,dots,n$ are "separated" (e.g. like $star7star5star4star6star$ if $n=7$) and $n-2$ slots/stars exists.



                          These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.



                          So we must select $3$ of $n-2$ stars giving factor $binom{n-2}3$.



                          Then we end up with a total of: $$3!(n-3)!binom{n-2}3$$possibilities.






                          share|cite|improve this answer


























                            0














                            There is a permutation of $1,2,3$ giving factor $3!$



                            There is a permutation of $4,5,dots n$ giving factor $(n-3)!$



                            The numbers $4,5,dots,n$ are "separated" (e.g. like $star7star5star4star6star$ if $n=7$) and $n-2$ slots/stars exists.



                            These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.



                            So we must select $3$ of $n-2$ stars giving factor $binom{n-2}3$.



                            Then we end up with a total of: $$3!(n-3)!binom{n-2}3$$possibilities.






                            share|cite|improve this answer
























                              0












                              0








                              0






                              There is a permutation of $1,2,3$ giving factor $3!$



                              There is a permutation of $4,5,dots n$ giving factor $(n-3)!$



                              The numbers $4,5,dots,n$ are "separated" (e.g. like $star7star5star4star6star$ if $n=7$) and $n-2$ slots/stars exists.



                              These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.



                              So we must select $3$ of $n-2$ stars giving factor $binom{n-2}3$.



                              Then we end up with a total of: $$3!(n-3)!binom{n-2}3$$possibilities.






                              share|cite|improve this answer












                              There is a permutation of $1,2,3$ giving factor $3!$



                              There is a permutation of $4,5,dots n$ giving factor $(n-3)!$



                              The numbers $4,5,dots,n$ are "separated" (e.g. like $star7star5star4star6star$ if $n=7$) and $n-2$ slots/stars exists.



                              These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.



                              So we must select $3$ of $n-2$ stars giving factor $binom{n-2}3$.



                              Then we end up with a total of: $$3!(n-3)!binom{n-2}3$$possibilities.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 1 at 10:25









                              drhab

                              97.4k544128




                              97.4k544128























                                  0














                                  We can count the cases




                                  • the permutations are n!


                                  from which we need to eliminate the ways




                                  • we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$

                                  • we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$






                                  share|cite|improve this answer




























                                    0














                                    We can count the cases




                                    • the permutations are n!


                                    from which we need to eliminate the ways




                                    • we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$

                                    • we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$






                                    share|cite|improve this answer


























                                      0












                                      0








                                      0






                                      We can count the cases




                                      • the permutations are n!


                                      from which we need to eliminate the ways




                                      • we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$

                                      • we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$






                                      share|cite|improve this answer














                                      We can count the cases




                                      • the permutations are n!


                                      from which we need to eliminate the ways




                                      • we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$

                                      • we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 1 at 11:05

























                                      answered Dec 1 at 10:13









                                      gimusi

                                      1




                                      1






























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