Binomial coefficients and words of $n$ bits, why this works?

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I read this on a webpage about the binomial coefficients:




If $0,1$ are bits and $k$ is the amount of "ones" $nchoosek$ gives the quantity of strings of length $n$ which have exactly $k$ ones in it.




Why is this true? I thought until now that the binomial coefficient $achooseb$ gave only informations about the numbers of ways you can choose a subsets of cardinality $b$ from one whose cardinality is $a$.




combinatorics




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  • Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
    – Steve Kass
    Aug 6 at 21:36










  • That was what I wanted to know! Thanks.
    – Baffo rasta
    Aug 6 at 21:37














up vote
1
down vote

favorite












I read this on a webpage about the binomial coefficients:




If $0,1$ are bits and $k$ is the amount of "ones" $nchoosek$ gives the quantity of strings of length $n$ which have exactly $k$ ones in it.




Why is this true? I thought until now that the binomial coefficient $achooseb$ gave only informations about the numbers of ways you can choose a subsets of cardinality $b$ from one whose cardinality is $a$.




combinatorics




share|cite|improve this question





















  • Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
    – Steve Kass
    Aug 6 at 21:36










  • That was what I wanted to know! Thanks.
    – Baffo rasta
    Aug 6 at 21:37












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I read this on a webpage about the binomial coefficients:




If $0,1$ are bits and $k$ is the amount of "ones" $nchoosek$ gives the quantity of strings of length $n$ which have exactly $k$ ones in it.




Why is this true? I thought until now that the binomial coefficient $achooseb$ gave only informations about the numbers of ways you can choose a subsets of cardinality $b$ from one whose cardinality is $a$.




combinatorics




share|cite|improve this question













I read this on a webpage about the binomial coefficients:




If $0,1$ are bits and $k$ is the amount of "ones" $nchoosek$ gives the quantity of strings of length $n$ which have exactly $k$ ones in it.




Why is this true? I thought until now that the binomial coefficient $achooseb$ gave only informations about the numbers of ways you can choose a subsets of cardinality $b$ from one whose cardinality is $a$.





combinatorics





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 21:26









Davide Morgante

1,930220




1,930220









asked Aug 6 at 21:23









Baffo rasta

15110




15110











  • Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
    – Steve Kass
    Aug 6 at 21:36










  • That was what I wanted to know! Thanks.
    – Baffo rasta
    Aug 6 at 21:37
















  • Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
    – Steve Kass
    Aug 6 at 21:36










  • That was what I wanted to know! Thanks.
    – Baffo rasta
    Aug 6 at 21:37















Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
– Steve Kass
Aug 6 at 21:36




Or more briefly than the answers (and like you asked in a comment): It works because there are $nchoose k$ ways to choose a set of $k$ positions to put the $1$s in from among the $n$ available positions in a length-$n$ string.
– Steve Kass
Aug 6 at 21:36












That was what I wanted to know! Thanks.
– Baffo rasta
Aug 6 at 21:37




That was what I wanted to know! Thanks.
– Baffo rasta
Aug 6 at 21:37










3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










Consider the following transformation: let $B$ be a set of cardinality $b$ and let us represent a subset $A$ of $B$ as a binary string of length $b$ with a one if the corresponding element is taken and a zero otherwise.



E.g.
$$B= u, v, w, x, y, z$$



$$A= u, v, zleftrightarrow 110001.$$



Now how many strings of length $b$ having $a$ ones are there ?






share|cite|improve this answer




























    up vote
    1
    down vote













    Let's rephrase the problem as the number of ways that a sequence of length $n$ can have $k$ ones.



    In some sense, this becomes a combinatorics problem since we are selecting $k$ positions out of the $n$ to be "ones", and the rest to be zeros.



    So, that explains why the result is $nchoosek$, since we are $textitchoosing$ $k$ ones from $n$ positions.






    share|cite|improve this answer

















    • 1




      My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
      – Baffo rasta
      Aug 6 at 21:30










    • Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
      – Rushabh Mehta
      Aug 6 at 21:32

















    up vote
    0
    down vote













    Here's a bijective argument, let $mathcalP_k([n])$ be the set of subsets of $1, dots, n$ of size $k$, and let



    $$
    S = (b_1, dots, b_n) in 0,1^n
    $$



    Now, the mapping



    $$
    Lambda: mathcalP_k([n]) longrightarrow S \
    A mapsto Lambda(A)
    $$



    with



    $$
    Lambda(A)_i := cases1 text if i in A\0 text if i not in A
    $$



    is bijective.



    In effect, if $A neq B$, wlog there exists $j in A$ such that $j not in B$. Thus, $Lambda(A)_j = 1 neq 0 = Lambda(B)_j$ and so $Lambda(A) neq Lambda(B)$ which proves that $Lambda$ is injective. Finally, if $b in S$, you can check that $A_b := j in [n] : b_j = 1$ has size $k$ and verifies $Lambda(A_b) = b$.






    share|cite|improve this answer























    • Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
      – Baffo rasta
      Aug 6 at 21:35










    • This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
      – Guido A.
      Aug 6 at 21:37










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Consider the following transformation: let $B$ be a set of cardinality $b$ and let us represent a subset $A$ of $B$ as a binary string of length $b$ with a one if the corresponding element is taken and a zero otherwise.



    E.g.
    $$B= u, v, w, x, y, z$$



    $$A= u, v, zleftrightarrow 110001.$$



    Now how many strings of length $b$ having $a$ ones are there ?






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Consider the following transformation: let $B$ be a set of cardinality $b$ and let us represent a subset $A$ of $B$ as a binary string of length $b$ with a one if the corresponding element is taken and a zero otherwise.



      E.g.
      $$B= u, v, w, x, y, z$$



      $$A= u, v, zleftrightarrow 110001.$$



      Now how many strings of length $b$ having $a$ ones are there ?






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Consider the following transformation: let $B$ be a set of cardinality $b$ and let us represent a subset $A$ of $B$ as a binary string of length $b$ with a one if the corresponding element is taken and a zero otherwise.



        E.g.
        $$B= u, v, w, x, y, z$$



        $$A= u, v, zleftrightarrow 110001.$$



        Now how many strings of length $b$ having $a$ ones are there ?






        share|cite|improve this answer













        Consider the following transformation: let $B$ be a set of cardinality $b$ and let us represent a subset $A$ of $B$ as a binary string of length $b$ with a one if the corresponding element is taken and a zero otherwise.



        E.g.
        $$B= u, v, w, x, y, z$$



        $$A= u, v, zleftrightarrow 110001.$$



        Now how many strings of length $b$ having $a$ ones are there ?







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 21:30









        Yves Daoust

        111k665205




        111k665205




















            up vote
            1
            down vote













            Let's rephrase the problem as the number of ways that a sequence of length $n$ can have $k$ ones.



            In some sense, this becomes a combinatorics problem since we are selecting $k$ positions out of the $n$ to be "ones", and the rest to be zeros.



            So, that explains why the result is $nchoosek$, since we are $textitchoosing$ $k$ ones from $n$ positions.






            share|cite|improve this answer

















            • 1




              My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
              – Baffo rasta
              Aug 6 at 21:30










            • Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
              – Rushabh Mehta
              Aug 6 at 21:32














            up vote
            1
            down vote













            Let's rephrase the problem as the number of ways that a sequence of length $n$ can have $k$ ones.



            In some sense, this becomes a combinatorics problem since we are selecting $k$ positions out of the $n$ to be "ones", and the rest to be zeros.



            So, that explains why the result is $nchoosek$, since we are $textitchoosing$ $k$ ones from $n$ positions.






            share|cite|improve this answer

















            • 1




              My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
              – Baffo rasta
              Aug 6 at 21:30










            • Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
              – Rushabh Mehta
              Aug 6 at 21:32












            up vote
            1
            down vote










            up vote
            1
            down vote









            Let's rephrase the problem as the number of ways that a sequence of length $n$ can have $k$ ones.



            In some sense, this becomes a combinatorics problem since we are selecting $k$ positions out of the $n$ to be "ones", and the rest to be zeros.



            So, that explains why the result is $nchoosek$, since we are $textitchoosing$ $k$ ones from $n$ positions.






            share|cite|improve this answer













            Let's rephrase the problem as the number of ways that a sequence of length $n$ can have $k$ ones.



            In some sense, this becomes a combinatorics problem since we are selecting $k$ positions out of the $n$ to be "ones", and the rest to be zeros.



            So, that explains why the result is $nchoosek$, since we are $textitchoosing$ $k$ ones from $n$ positions.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Aug 6 at 21:27









            Rushabh Mehta

            1,060114




            1,060114







            • 1




              My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
              – Baffo rasta
              Aug 6 at 21:30










            • Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
              – Rushabh Mehta
              Aug 6 at 21:32












            • 1




              My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
              – Baffo rasta
              Aug 6 at 21:30










            • Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
              – Rushabh Mehta
              Aug 6 at 21:32







            1




            1




            My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
            – Baffo rasta
            Aug 6 at 21:30




            My initial thought was that in this case $nchoosek$ could be giving me the number of different groups of indexes that could be taken by the 1 digits in a single string of $n$ bits. Is this correct?
            – Baffo rasta
            Aug 6 at 21:30












            Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
            – Rushabh Mehta
            Aug 6 at 21:32




            Well, not exactly. $nchoose k$ gives you the number of ways to pick $k$ things from a group of $n$ things. In this case, you are picking $k$ places out of $n$ places to put 1s, and the rest will be assigned 0s.
            – Rushabh Mehta
            Aug 6 at 21:32










            up vote
            0
            down vote













            Here's a bijective argument, let $mathcalP_k([n])$ be the set of subsets of $1, dots, n$ of size $k$, and let



            $$
            S = (b_1, dots, b_n) in 0,1^n
            $$



            Now, the mapping



            $$
            Lambda: mathcalP_k([n]) longrightarrow S \
            A mapsto Lambda(A)
            $$



            with



            $$
            Lambda(A)_i := cases1 text if i in A\0 text if i not in A
            $$



            is bijective.



            In effect, if $A neq B$, wlog there exists $j in A$ such that $j not in B$. Thus, $Lambda(A)_j = 1 neq 0 = Lambda(B)_j$ and so $Lambda(A) neq Lambda(B)$ which proves that $Lambda$ is injective. Finally, if $b in S$, you can check that $A_b := j in [n] : b_j = 1$ has size $k$ and verifies $Lambda(A_b) = b$.






            share|cite|improve this answer























            • Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
              – Baffo rasta
              Aug 6 at 21:35










            • This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
              – Guido A.
              Aug 6 at 21:37














            up vote
            0
            down vote













            Here's a bijective argument, let $mathcalP_k([n])$ be the set of subsets of $1, dots, n$ of size $k$, and let



            $$
            S = (b_1, dots, b_n) in 0,1^n
            $$



            Now, the mapping



            $$
            Lambda: mathcalP_k([n]) longrightarrow S \
            A mapsto Lambda(A)
            $$



            with



            $$
            Lambda(A)_i := cases1 text if i in A\0 text if i not in A
            $$



            is bijective.



            In effect, if $A neq B$, wlog there exists $j in A$ such that $j not in B$. Thus, $Lambda(A)_j = 1 neq 0 = Lambda(B)_j$ and so $Lambda(A) neq Lambda(B)$ which proves that $Lambda$ is injective. Finally, if $b in S$, you can check that $A_b := j in [n] : b_j = 1$ has size $k$ and verifies $Lambda(A_b) = b$.






            share|cite|improve this answer























            • Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
              – Baffo rasta
              Aug 6 at 21:35










            • This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
              – Guido A.
              Aug 6 at 21:37












            up vote
            0
            down vote










            up vote
            0
            down vote









            Here's a bijective argument, let $mathcalP_k([n])$ be the set of subsets of $1, dots, n$ of size $k$, and let



            $$
            S = (b_1, dots, b_n) in 0,1^n
            $$



            Now, the mapping



            $$
            Lambda: mathcalP_k([n]) longrightarrow S \
            A mapsto Lambda(A)
            $$



            with



            $$
            Lambda(A)_i := cases1 text if i in A\0 text if i not in A
            $$



            is bijective.



            In effect, if $A neq B$, wlog there exists $j in A$ such that $j not in B$. Thus, $Lambda(A)_j = 1 neq 0 = Lambda(B)_j$ and so $Lambda(A) neq Lambda(B)$ which proves that $Lambda$ is injective. Finally, if $b in S$, you can check that $A_b := j in [n] : b_j = 1$ has size $k$ and verifies $Lambda(A_b) = b$.






            share|cite|improve this answer















            Here's a bijective argument, let $mathcalP_k([n])$ be the set of subsets of $1, dots, n$ of size $k$, and let



            $$
            S = (b_1, dots, b_n) in 0,1^n
            $$



            Now, the mapping



            $$
            Lambda: mathcalP_k([n]) longrightarrow S \
            A mapsto Lambda(A)
            $$



            with



            $$
            Lambda(A)_i := cases1 text if i in A\0 text if i not in A
            $$



            is bijective.



            In effect, if $A neq B$, wlog there exists $j in A$ such that $j not in B$. Thus, $Lambda(A)_j = 1 neq 0 = Lambda(B)_j$ and so $Lambda(A) neq Lambda(B)$ which proves that $Lambda$ is injective. Finally, if $b in S$, you can check that $A_b := j in [n] : b_j = 1$ has size $k$ and verifies $Lambda(A_b) = b$.







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 6 at 21:40


























            answered Aug 6 at 21:33









            Guido A.

            3,851624




            3,851624











            • Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
              – Baffo rasta
              Aug 6 at 21:35










            • This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
              – Guido A.
              Aug 6 at 21:37
















            • Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
              – Baffo rasta
              Aug 6 at 21:35










            • This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
              – Guido A.
              Aug 6 at 21:37















            Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
            – Baffo rasta
            Aug 6 at 21:35




            Well, I kind of get if this is the case what I am missing! That one is the argument of the next chapter of my book!
            – Baffo rasta
            Aug 6 at 21:35












            This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
            – Guido A.
            Aug 6 at 21:37




            This is a very common type of 'combinatorial proof'. Say you want to prove that $n = m$ for some natural numbers $n,m$; then sometimes these can be interpreted as the sizes of certain sets. Thus, $n = m$ will happen iff there is a bijection between these. That motivates this type of arguments. I'll add a proof of my claim for completion's sake :)
            – Guido A.
            Aug 6 at 21:37












             

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            Color the edges and diagonals of a regular polygon

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            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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