Number of ways to pick elements such mutual ordering remains maintained?












-1












$begingroup$


The number of ways to pick elements from two sets such that mutual ordering remains intact of both the sets.



For example:

Set 1:(a,b)

Set2:(c,d)

What I mean is a should come before b in any order you pick similarly c before d.

Number of ways is 6

{abcd,acdb,acbd,cabd,cadb,cdab}





combinatorics permutations







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$endgroup$












  • $begingroup$
    Hint: Stars_and_bars.
    $endgroup$
    – drhab
    Dec 30 '18 at 8:11










  • $begingroup$
    Got it!! Thanks!!
    $endgroup$
    – Sagar Sharma
    Dec 30 '18 at 8:20










  • $begingroup$
    @bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:04












  • $begingroup$
    @bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:15


















-1












$begingroup$


The number of ways to pick elements from two sets such that mutual ordering remains intact of both the sets.



For example:

Set 1:(a,b)

Set2:(c,d)

What I mean is a should come before b in any order you pick similarly c before d.

Number of ways is 6

{abcd,acdb,acbd,cabd,cadb,cdab}





combinatorics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Stars_and_bars.
    $endgroup$
    – drhab
    Dec 30 '18 at 8:11










  • $begingroup$
    Got it!! Thanks!!
    $endgroup$
    – Sagar Sharma
    Dec 30 '18 at 8:20










  • $begingroup$
    @bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:04












  • $begingroup$
    @bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:15
















-1












-1








-1





$begingroup$


The number of ways to pick elements from two sets such that mutual ordering remains intact of both the sets.



For example:

Set 1:(a,b)

Set2:(c,d)

What I mean is a should come before b in any order you pick similarly c before d.

Number of ways is 6

{abcd,acdb,acbd,cabd,cadb,cdab}





combinatorics permutations







share|cite|improve this question











$endgroup$




The number of ways to pick elements from two sets such that mutual ordering remains intact of both the sets.



For example:

Set 1:(a,b)

Set2:(c,d)

What I mean is a should come before b in any order you pick similarly c before d.

Number of ways is 6

{abcd,acdb,acbd,cabd,cadb,cdab}





combinatorics permutations




combinatorics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 7:57









Asaf Karagila♦

307k33438769




307k33438769










asked Dec 30 '18 at 7:52









Sagar SharmaSagar Sharma

143




143












  • $begingroup$
    Hint: Stars_and_bars.
    $endgroup$
    – drhab
    Dec 30 '18 at 8:11










  • $begingroup$
    Got it!! Thanks!!
    $endgroup$
    – Sagar Sharma
    Dec 30 '18 at 8:20










  • $begingroup$
    @bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:04












  • $begingroup$
    @bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:15




















  • $begingroup$
    Hint: Stars_and_bars.
    $endgroup$
    – drhab
    Dec 30 '18 at 8:11










  • $begingroup$
    Got it!! Thanks!!
    $endgroup$
    – Sagar Sharma
    Dec 30 '18 at 8:20










  • $begingroup$
    @bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:04












  • $begingroup$
    @bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
    $endgroup$
    – drhab
    Dec 30 '18 at 9:15


















$begingroup$
Hint: Stars_and_bars.
$endgroup$
– drhab
Dec 30 '18 at 8:11




$begingroup$
Hint: Stars_and_bars.
$endgroup$
– drhab
Dec 30 '18 at 8:11












$begingroup$
Got it!! Thanks!!
$endgroup$
– Sagar Sharma
Dec 30 '18 at 8:20




$begingroup$
Got it!! Thanks!!
$endgroup$
– Sagar Sharma
Dec 30 '18 at 8:20












$begingroup$
@bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
$endgroup$
– drhab
Dec 30 '18 at 9:04






$begingroup$
@bof Suppose that the (ordered) sets have $n$ and $m$ elements. Then to be found is the number of sums $k_1+cdots+k_{n+1}=m$ where the $k_i$ are nonnegative integers. First place the elements of the set having $n$ elements on a row. Then place elements of the other set in the gaps and on LHS and RHS.
$endgroup$
– drhab
Dec 30 '18 at 9:04














$begingroup$
@bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
$endgroup$
– drhab
Dec 30 '18 at 9:15






$begingroup$
@bof I must agree it is more trouble :-). It was the first thing that came to mind and was projected immediately as a hint. Fortunately it was well-received by the OP.
$endgroup$
– drhab
Dec 30 '18 at 9:15












1 Answer
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oldest

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1












$begingroup$

In general, if the first ordered set has $m$ elements and the second $n$, then every such order is found by having $n+m$ places reserved and picking the $m$ places where the elements of the first set are going to go (in order). The remaining places get the second set elements (also in order).



Picking $m$ places from $n+m$ can be done in $binom{n+m}{m}$ ways. Note that we could also pick the places for set $2$ first too, but this gives the same result as $binom{n+m}{n} = binom{n+m}{m}$.



In your case $n=m=2$ and indeed $binom{4}{2} = frac{4!}{2! 2!} = frac{24}{4}=6$.






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    1 Answer
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    $begingroup$

    In general, if the first ordered set has $m$ elements and the second $n$, then every such order is found by having $n+m$ places reserved and picking the $m$ places where the elements of the first set are going to go (in order). The remaining places get the second set elements (also in order).



    Picking $m$ places from $n+m$ can be done in $binom{n+m}{m}$ ways. Note that we could also pick the places for set $2$ first too, but this gives the same result as $binom{n+m}{n} = binom{n+m}{m}$.



    In your case $n=m=2$ and indeed $binom{4}{2} = frac{4!}{2! 2!} = frac{24}{4}=6$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      In general, if the first ordered set has $m$ elements and the second $n$, then every such order is found by having $n+m$ places reserved and picking the $m$ places where the elements of the first set are going to go (in order). The remaining places get the second set elements (also in order).



      Picking $m$ places from $n+m$ can be done in $binom{n+m}{m}$ ways. Note that we could also pick the places for set $2$ first too, but this gives the same result as $binom{n+m}{n} = binom{n+m}{m}$.



      In your case $n=m=2$ and indeed $binom{4}{2} = frac{4!}{2! 2!} = frac{24}{4}=6$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        In general, if the first ordered set has $m$ elements and the second $n$, then every such order is found by having $n+m$ places reserved and picking the $m$ places where the elements of the first set are going to go (in order). The remaining places get the second set elements (also in order).



        Picking $m$ places from $n+m$ can be done in $binom{n+m}{m}$ ways. Note that we could also pick the places for set $2$ first too, but this gives the same result as $binom{n+m}{n} = binom{n+m}{m}$.



        In your case $n=m=2$ and indeed $binom{4}{2} = frac{4!}{2! 2!} = frac{24}{4}=6$.






        share|cite|improve this answer











        $endgroup$



        In general, if the first ordered set has $m$ elements and the second $n$, then every such order is found by having $n+m$ places reserved and picking the $m$ places where the elements of the first set are going to go (in order). The remaining places get the second set elements (also in order).



        Picking $m$ places from $n+m$ can be done in $binom{n+m}{m}$ ways. Note that we could also pick the places for set $2$ first too, but this gives the same result as $binom{n+m}{n} = binom{n+m}{m}$.



        In your case $n=m=2$ and indeed $binom{4}{2} = frac{4!}{2! 2!} = frac{24}{4}=6$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 9:30

























        answered Dec 30 '18 at 9:19









        Henno BrandsmaHenno Brandsma

        113k348123




        113k348123






























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